# Thread: Trigonometric identity (solution needed)

1. ## Trigonometric identity (solution needed)

sin(a+b)2 + cos(a+b)2 = 1

Can the above be solved in a way that the terms with "a" and with "b" are separated. It does not matter that in the process sin and cos are replaced with the other functions (tan, cot, csc or sec). Any help would be greatly appreciated. Thanks.

2. ## Re: Trigonometric identity (solution needed)

Ambiguous: do you mean $sin^2(a+ b)$ or $sin((a+b)^2)$? In either case, why "separate" a and b? Let $\theta= a+ b$.

3. ## Re: Trigonometric identity (solution needed)

sin(a+b)2 + cos(a+b)2 = 1

Can the above be solved in a way that the terms with "a" and with "b" are separated. It does not matter that in the process sin and cos are replaced with the other functions (tan, cot, csc or sec). Any help would be greatly appreciated. Thanks.
Are you perhaps looking for a solution for b in terms of a? What you have written is an identity...it works for any values of a and b.

-Dan

4. ## Re: Trigonometric identity (solution needed)

I meant the first one. What I am trying to achieve is to separate the a and b and later on after calculating them separately add them the results to get back to the same identity. Let me explain with an example. Suppose a is 50 and b is 30. Now what I want is a solution that will give me something like (sin square of 50 + sin square of 30) + (cos square of 50 + cos square of 30) still equals 1. I don't know if you got my point or not, but my target is to separate the two angles in any way, don't dwell too much on the example, so that I can separate the a and b and still have the result 1.

5. ## Re: Trigonometric identity (solution needed)

No, I want to keep the a and b separate.

6. ## Re: Trigonometric identity (solution needed)

I meant the first one.
Which are you calling the first?
No, I want to keep the a and b separate.
If you mean $\large{\sin^2(a+b)+\cos^2(a+b)=1}$

Then that equation is satisfied for any value of $\large{\bf{a}\text{ or }\bf{b}\text{ whatsoever }}$

7. ## Re: Trigonometric identity (solution needed)

sin(a+b)2 + cos(a+b)2 = 1

Can the above be solved in a way that the terms with "a" and with "b" are separated. It does not matter that in the process sin and cos are replaced with the other functions (tan, cot, csc or sec). Any help would be greatly appreciated. Thanks.
I'm not sure what you mean by "solved". Maybe you mean "proven" using the identities $sin^2{a}+cos^2{a} =1$ and $sin^2{b}+cos^2{b} =1$ ??

If this is what you mean then maybe this will help:

$sin^2(a+b)+cos^2(a+b)$

$= (sin(a+b))^2 + (cos(a+b))^2$

$= (sin a cos b + sin b cos a)^2 + (cos a cos b - sin a sin b)^2$

$= sin^2 a cos^2 b + 2 sin a cos b sin b cos a + sin^2 b cos^2 a + cos^2 a cos^2 b - 2 cos a cos b sin a sin b + sin^2 a sin^2 b$

$= sin^2 a cos^2 b + sin^2 b cos^2 a + cos^2 a cos^2 b + sin^2 a sin^2 b$

$= sin^2 a cos^2 b + + sin^2 a sin^2 b +sin^2 b cos^2 a + cos^2 a cos^2 b$

$= sin^2 a (cos^2 b + sin^2 b) + cos^2 a(sin^2 b + cos^2 b)$

$= sin^2 a + cos^2 a$

$=1$

8. ## Re: Trigonometric identity (solution needed)

Maybe I am not explaining it properly. Let me try again.

Suppose I have broken the theta in a and b and I give 'a' to someone and say that you can calculate an equation (having sin, cos, tan ... etc.) with that 'a' and return the result to me. Then I calculate the same equation using 'b' myself and add the results to get 1. So basically it will be something of this sort

sin ^2 (a+b) + cos ^2 (a+b) = some_equation_with_a_only + some_equation_with_b_only = 1

Hope that makes it clear :-)

9. ## Re: Trigonometric identity (solution needed)

Then you still have not understood what every one is trying to tell you. $sin^2(x)+ cos^2(x)= 1$ for all x. Writing x as a+ b does not change that! $sin^2(a+ b)+ cos^2(a+ b)= 1$ for all possible values of a and b.

10. ## Re: Trigonometric identity (solution needed)

Thank you all firstly for the prompt replies. If I have sin^2(ab) + cos^2(ab) = 1, how will I expand this ?

11. ## Re: Trigonometric identity (solution needed)

I don't think you understand the word IDENTITY ...

$\sin^2(\text{whatever}) + \cos^2(\text{the same whatever}) = 1$

... in other words, the quantity $(ab)$ can be whatever value you want it to be.

12. ## Re: Trigonometric identity (solution needed)

Let Sin (a+b)2 + cos (a+b)2 = 1
Let (a+b)2 = k, then we have
Sin k + cos k = 1
Now proceed to solve this equation