challenge trig question - no calculator
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2
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The challenge trignometry question is: simplify $sin (80^circ) + sin (40^circ) $ using trignometric identities. All the angle values are in degrees.
This is what I did: Let $a=40^circ$. So we have $sin(2a) +sin(a)$. Using the formula for $sin(2a)$, I have,
$$2sin(a)cos(a) + sin(a)=sin(40)(2cos(40^circ)+1)$$
All I got was a big red mark through the question. I looked at co-functions and half-angles and couldn't find a solution that came out to something I could do without a calculator. Any thoughts?
algebra-precalculus trigonometry
edited Nov 13 at 23:36
User525412790
313114
asked Nov 13 at 22:42
user163862
7761916
add a comment |
up vote
2
down vote
favorite
The challenge trignometry question is: simplify $sin (80^circ) + sin (40^circ) $ using trignometric identities. All the angle values are in degrees.
This is what I did: Let $a=40^circ$. So we have $sin(2a) +sin(a)$. Using the formula for $sin(2a)$, I have,
$$2sin(a)cos(a) + sin(a)=sin(40)(2cos(40^circ)+1)$$
All I got was a big red mark through the question. I looked at co-functions and half-angles and couldn't find a solution that came out to something I could do without a calculator. Any thoughts?
algebra-precalculus trigonometry
edited Nov 13 at 23:36
User525412790
313114
asked Nov 13 at 22:42
user163862
7761916
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so sin x gives $sin x$
– Ross Millikan
Nov 13 at 22:52
1
A 'big red mark' should urge you to ask for the solution from your instructor. They will only be too happy to teach!
– User525412790
Nov 13 at 22:59
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The challenge trignometry question is: simplify $sin (80^circ) + sin (40^circ) $ using trignometric identities. All the angle values are in degrees.
This is what I did: Let $a=40^circ$. So we have $sin(2a) +sin(a)$. Using the formula for $sin(2a)$, I have,
$$2sin(a)cos(a) + sin(a)=sin(40)(2cos(40^circ)+1)$$
All I got was a big red mark through the question. I looked at co-functions and half-angles and couldn't find a solution that came out to something I could do without a calculator. Any thoughts?
algebra-precalculus trigonometry
edited Nov 13 at 23:36
User525412790
313114
asked Nov 13 at 22:42
user163862
7761916
The challenge trignometry question is: simplify $sin (80^circ) + sin (40^circ) $ using trignometric identities. All the angle values are in degrees.
This is what I did: Let $a=40^circ$. So we have $sin(2a) +sin(a)$. Using the formula for $sin(2a)$, I have,
$$2sin(a)cos(a) + sin(a)=sin(40)(2cos(40^circ)+1)$$
All I got was a big red mark through the question. I looked at co-functions and half-angles and couldn't find a solution that came out to something I could do without a calculator. Any thoughts?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Nov 13 at 23:36
User525412790
313114
asked Nov 13 at 22:42
user163862
7761916
edited Nov 13 at 23:36
User525412790
313114
asked Nov 13 at 22:42
user163862
7761916
edited Nov 13 at 23:36
User525412790
313114
edited Nov 13 at 23:36
User525412790
313114
edited Nov 13 at 23:36
User525412790
313114
313114
asked Nov 13 at 22:42
user163862
7761916
asked Nov 13 at 22:42
user163862
7761916
asked Nov 13 at 22:42
user163862
7761916
7761916
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so sin x gives $sin x$
– Ross Millikan
Nov 13 at 22:52
1
A 'big red mark' should urge you to ask for the solution from your instructor. They will only be too happy to teach!
– User525412790
Nov 13 at 22:59
add a comment |
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so sin x gives $sin x$
– Ross Millikan
Nov 13 at 22:52
1
A 'big red mark' should urge you to ask for the solution from your instructor. They will only be too happy to teach!
– User525412790
Nov 13 at 22:59
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so sin x gives $sin x$
– Ross Millikan
Nov 13 at 22:52
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so sin x gives $sin x$
– Ross Millikan
Nov 13 at 22:52
1
1
A 'big red mark' should urge you to ask for the solution from your instructor. They will only be too happy to teach!
– User525412790
Nov 13 at 22:59
A 'big red mark' should urge you to ask for the solution from your instructor. They will only be too happy to teach!
– User525412790
Nov 13 at 22:59
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
You may have been intended to write
$$begin {align}sin (80)+sin (40)&=2sinleft(frac {80+40}2right)cosleft(frac {80-40}2right)\&=2sin (60) cos (20)\&=sqrt 3 cos (20) end {align}$$
but I don't have a nice value for $cos(20)$ and neither does Alpha, which finds about $0.93969$. The first two terms of the Taylor series should be close, and if you know $pi^2approx 10$ you can write $cos(20)approx 1-frac 12(frac {pi}9)^2approx1-frac {10}{162}approx 1-frac 1{16}=0.9375$ which is very close.
answered Nov 13 at 23:00
Ross Millikan
287k23195364
I think that you are right and $sqrt 3 cos (20)$ was the requested result.
– gimusi
Nov 13 at 23:05
Thank you. I didn't recall that formula.
– user163862
Nov 13 at 23:13
@user163862: I had to look it up. I know the angle-sum formulas, but not the function-sum ones.
– Ross Millikan
Nov 13 at 23:15
After examining that formula, I can see that $sin (A+B) + Sin (A-B)=2Sin(A)Cos(B)$. But it would seem to me that when I divide by $2$ it divides the $sin$ term not the angles
– user163862
Nov 14 at 20:24
If you substitute $A=60,B=20$ into your formula you get mine exactly. This comes from solving $A+B=80,A-B=40$ and the divide by $2$ comes when you add and subtract the two equations.
– Ross Millikan
Nov 14 at 20:53
add a comment |
up vote
2
down vote
Maybe we need to a single angle, in that case we have that
- ${{sin(theta + varphi) + sin(theta - varphi)} }=2sin theta
cos varphi$
therefore
$$sin (80)+sin (40) = 2sin (60)cos (20)=sqrt 3 cos 20$$
answered Nov 13 at 23:00
gimusi
86.9k74393
add a comment |
up vote
1
down vote
You're supposed to find a numerical value.
sin 3x = 3.sin x - 4.sin$^4$ x
will give values for sin 40 deg.
answered Nov 13 at 23:08
William Elliot
6,7902518
Actually the problem said to simplify; however, I'd be very interested in getting the numerical value without a calculator. How do I do that using the sin (3x) formula.
– user163862
Nov 14 at 20:08
@user163862. What is sin 3×40 degrees?
– William Elliot
Nov 15 at 4:14
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You may have been intended to write
$$begin {align}sin (80)+sin (40)&=2sinleft(frac {80+40}2right)cosleft(frac {80-40}2right)\&=2sin (60) cos (20)\&=sqrt 3 cos (20) end {align}$$
but I don't have a nice value for $cos(20)$ and neither does Alpha, which finds about $0.93969$. The first two terms of the Taylor series should be close, and if you know $pi^2approx 10$ you can write $cos(20)approx 1-frac 12(frac {pi}9)^2approx1-frac {10}{162}approx 1-frac 1{16}=0.9375$ which is very close.
answered Nov 13 at 23:00
Ross Millikan
287k23195364
I think that you are right and $sqrt 3 cos (20)$ was the requested result.
– gimusi
Nov 13 at 23:05
Thank you. I didn't recall that formula.
– user163862
Nov 13 at 23:13
@user163862: I had to look it up. I know the angle-sum formulas, but not the function-sum ones.
– Ross Millikan
Nov 13 at 23:15
After examining that formula, I can see that $sin (A+B) + Sin (A-B)=2Sin(A)Cos(B)$. But it would seem to me that when I divide by $2$ it divides the $sin$ term not the angles
– user163862
Nov 14 at 20:24
If you substitute $A=60,B=20$ into your formula you get mine exactly. This comes from solving $A+B=80,A-B=40$ and the divide by $2$ comes when you add and subtract the two equations.
– Ross Millikan
Nov 14 at 20:53
add a comment |
up vote
4
down vote
accepted
You may have been intended to write
$$begin {align}sin (80)+sin (40)&=2sinleft(frac {80+40}2right)cosleft(frac {80-40}2right)\&=2sin (60) cos (20)\&=sqrt 3 cos (20) end {align}$$
but I don't have a nice value for $cos(20)$ and neither does Alpha, which finds about $0.93969$. The first two terms of the Taylor series should be close, and if you know $pi^2approx 10$ you can write $cos(20)approx 1-frac 12(frac {pi}9)^2approx1-frac {10}{162}approx 1-frac 1{16}=0.9375$ which is very close.
answered Nov 13 at 23:00
Ross Millikan
287k23195364
I think that you are right and $sqrt 3 cos (20)$ was the requested result.
– gimusi
Nov 13 at 23:05
Thank you. I didn't recall that formula.
– user163862
Nov 13 at 23:13
@user163862: I had to look it up. I know the angle-sum formulas, but not the function-sum ones.
– Ross Millikan
Nov 13 at 23:15
After examining that formula, I can see that $sin (A+B) + Sin (A-B)=2Sin(A)Cos(B)$. But it would seem to me that when I divide by $2$ it divides the $sin$ term not the angles
– user163862
Nov 14 at 20:24
If you substitute $A=60,B=20$ into your formula you get mine exactly. This comes from solving $A+B=80,A-B=40$ and the divide by $2$ comes when you add and subtract the two equations.
– Ross Millikan
Nov 14 at 20:53
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You may have been intended to write
$$begin {align}sin (80)+sin (40)&=2sinleft(frac {80+40}2right)cosleft(frac {80-40}2right)\&=2sin (60) cos (20)\&=sqrt 3 cos (20) end {align}$$
but I don't have a nice value for $cos(20)$ and neither does Alpha, which finds about $0.93969$. The first two terms of the Taylor series should be close, and if you know $pi^2approx 10$ you can write $cos(20)approx 1-frac 12(frac {pi}9)^2approx1-frac {10}{162}approx 1-frac 1{16}=0.9375$ which is very close.
answered Nov 13 at 23:00
Ross Millikan
287k23195364
You may have been intended to write
$$begin {align}sin (80)+sin (40)&=2sinleft(frac {80+40}2right)cosleft(frac {80-40}2right)\&=2sin (60) cos (20)\&=sqrt 3 cos (20) end {align}$$
but I don't have a nice value for $cos(20)$ and neither does Alpha, which finds about $0.93969$. The first two terms of the Taylor series should be close, and if you know $pi^2approx 10$ you can write $cos(20)approx 1-frac 12(frac {pi}9)^2approx1-frac {10}{162}approx 1-frac 1{16}=0.9375$ which is very close.
answered Nov 13 at 23:00
Ross Millikan
287k23195364
answered Nov 13 at 23:00
Ross Millikan
287k23195364
answered Nov 13 at 23:00
Ross Millikan
287k23195364
answered Nov 13 at 23:00
Ross Millikan
287k23195364
287k23195364
I think that you are right and $sqrt 3 cos (20)$ was the requested result.
– gimusi
Nov 13 at 23:05
Thank you. I didn't recall that formula.
– user163862
Nov 13 at 23:13
@user163862: I had to look it up. I know the angle-sum formulas, but not the function-sum ones.
– Ross Millikan
Nov 13 at 23:15
After examining that formula, I can see that $sin (A+B) + Sin (A-B)=2Sin(A)Cos(B)$. But it would seem to me that when I divide by $2$ it divides the $sin$ term not the angles
– user163862
Nov 14 at 20:24
If you substitute $A=60,B=20$ into your formula you get mine exactly. This comes from solving $A+B=80,A-B=40$ and the divide by $2$ comes when you add and subtract the two equations.
– Ross Millikan
Nov 14 at 20:53
add a comment |
I think that you are right and $sqrt 3 cos (20)$ was the requested result.
– gimusi
Nov 13 at 23:05
Thank you. I didn't recall that formula.
– user163862
Nov 13 at 23:13
@user163862: I had to look it up. I know the angle-sum formulas, but not the function-sum ones.
– Ross Millikan
Nov 13 at 23:15
After examining that formula, I can see that $sin (A+B) + Sin (A-B)=2Sin(A)Cos(B)$. But it would seem to me that when I divide by $2$ it divides the $sin$ term not the angles
– user163862
Nov 14 at 20:24
If you substitute $A=60,B=20$ into your formula you get mine exactly. This comes from solving $A+B=80,A-B=40$ and the divide by $2$ comes when you add and subtract the two equations.
– Ross Millikan
Nov 14 at 20:53
I think that you are right and $sqrt 3 cos (20)$ was the requested result.
– gimusi
Nov 13 at 23:05
I think that you are right and $sqrt 3 cos (20)$ was the requested result.
– gimusi
Nov 13 at 23:05
Thank you. I didn't recall that formula.
– user163862
Nov 13 at 23:13
Thank you. I didn't recall that formula.
– user163862
Nov 13 at 23:13
@user163862: I had to look it up. I know the angle-sum formulas, but not the function-sum ones.
– Ross Millikan
Nov 13 at 23:15
@user163862: I had to look it up. I know the angle-sum formulas, but not the function-sum ones.
– Ross Millikan
Nov 13 at 23:15
After examining that formula, I can see that $sin (A+B) + Sin (A-B)=2Sin(A)Cos(B)$. But it would seem to me that when I divide by $2$ it divides the $sin$ term not the angles
– user163862
Nov 14 at 20:24
After examining that formula, I can see that $sin (A+B) + Sin (A-B)=2Sin(A)Cos(B)$. But it would seem to me that when I divide by $2$ it divides the $sin$ term not the angles
– user163862
Nov 14 at 20:24
If you substitute $A=60,B=20$ into your formula you get mine exactly. This comes from solving $A+B=80,A-B=40$ and the divide by $2$ comes when you add and subtract the two equations.
– Ross Millikan
Nov 14 at 20:53
If you substitute $A=60,B=20$ into your formula you get mine exactly. This comes from solving $A+B=80,A-B=40$ and the divide by $2$ comes when you add and subtract the two equations.
– Ross Millikan
Nov 14 at 20:53
add a comment |
up vote
2
down vote
Maybe we need to a single angle, in that case we have that
- ${{sin(theta + varphi) + sin(theta - varphi)} }=2sin theta
cos varphi$
therefore
$$sin (80)+sin (40) = 2sin (60)cos (20)=sqrt 3 cos 20$$
answered Nov 13 at 23:00
gimusi
86.9k74393
add a comment |
up vote
2
down vote
Maybe we need to a single angle, in that case we have that
- ${{sin(theta + varphi) + sin(theta - varphi)} }=2sin theta
cos varphi$
therefore
$$sin (80)+sin (40) = 2sin (60)cos (20)=sqrt 3 cos 20$$
answered Nov 13 at 23:00
gimusi
86.9k74393
add a comment |
up vote
2
down vote
up vote
2
down vote
Maybe we need to a single angle, in that case we have that
- ${{sin(theta + varphi) + sin(theta - varphi)} }=2sin theta
cos varphi$
therefore
$$sin (80)+sin (40) = 2sin (60)cos (20)=sqrt 3 cos 20$$
answered Nov 13 at 23:00
gimusi
86.9k74393
Maybe we need to a single angle, in that case we have that
- ${{sin(theta + varphi) + sin(theta - varphi)} }=2sin theta
cos varphi$
therefore
$$sin (80)+sin (40) = 2sin (60)cos (20)=sqrt 3 cos 20$$
answered Nov 13 at 23:00
gimusi
86.9k74393
answered Nov 13 at 23:00
gimusi
86.9k74393
answered Nov 13 at 23:00
gimusi
86.9k74393
answered Nov 13 at 23:00
gimusi
86.9k74393
86.9k74393
add a comment |
add a comment |
up vote
1
down vote
You're supposed to find a numerical value.
sin 3x = 3.sin x - 4.sin$^4$ x
will give values for sin 40 deg.
answered Nov 13 at 23:08
William Elliot
6,7902518
Actually the problem said to simplify; however, I'd be very interested in getting the numerical value without a calculator. How do I do that using the sin (3x) formula.
– user163862
Nov 14 at 20:08
@user163862. What is sin 3×40 degrees?
– William Elliot
Nov 15 at 4:14
add a comment |
up vote
1
down vote
You're supposed to find a numerical value.
sin 3x = 3.sin x - 4.sin$^4$ x
will give values for sin 40 deg.
answered Nov 13 at 23:08
William Elliot
6,7902518
Actually the problem said to simplify; however, I'd be very interested in getting the numerical value without a calculator. How do I do that using the sin (3x) formula.
– user163862
Nov 14 at 20:08
@user163862. What is sin 3×40 degrees?
– William Elliot
Nov 15 at 4:14
add a comment |
up vote
1
down vote
up vote
1
down vote
You're supposed to find a numerical value.
sin 3x = 3.sin x - 4.sin$^4$ x
will give values for sin 40 deg.
answered Nov 13 at 23:08
William Elliot
6,7902518
You're supposed to find a numerical value.
sin 3x = 3.sin x - 4.sin$^4$ x
will give values for sin 40 deg.
answered Nov 13 at 23:08
William Elliot
6,7902518
answered Nov 13 at 23:08
William Elliot
6,7902518
answered Nov 13 at 23:08
William Elliot
6,7902518
answered Nov 13 at 23:08
William Elliot
6,7902518
6,7902518
Actually the problem said to simplify; however, I'd be very interested in getting the numerical value without a calculator. How do I do that using the sin (3x) formula.
– user163862
Nov 14 at 20:08
@user163862. What is sin 3×40 degrees?
– William Elliot
Nov 15 at 4:14
add a comment |
Actually the problem said to simplify; however, I'd be very interested in getting the numerical value without a calculator. How do I do that using the sin (3x) formula.
– user163862
Nov 14 at 20:08
@user163862. What is sin 3×40 degrees?
– William Elliot
Nov 15 at 4:14
Actually the problem said to simplify; however, I'd be very interested in getting the numerical value without a calculator. How do I do that using the sin (3x) formula.
– user163862
Nov 14 at 20:08
Actually the problem said to simplify; however, I'd be very interested in getting the numerical value without a calculator. How do I do that using the sin (3x) formula.
– user163862
Nov 14 at 20:08
@user163862. What is sin 3×40 degrees?
– William Elliot
Nov 15 at 4:14
@user163862. What is sin 3×40 degrees?
– William Elliot
Nov 15 at 4:14
add a comment |
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MathJax hint: if you put a backslash before common functions you get the right font and spacing, so sin x gives $sin x$
– Ross Millikan
Nov 13 at 22:52
1
A 'big red mark' should urge you to ask for the solution from your instructor. They will only be too happy to teach!
– User525412790
Nov 13 at 22:59