# [SOLVED] Trigonometric problem

• March 16th 2010, 07:35 PM
Solaris123
[SOLVED] Trigonometric problem
Please tell me the solution and the steps required to get solution to this.

Let sin(a)= 1/ sqrt(5) 0 < a < 90 degrees
cos(b)= 3/ sqrt(10) 0< b <90

then what is sin(a+b) equal to?
• March 16th 2010, 07:41 PM
Anonymous1
Take the arcsine and arccosine to find a and b.
Then plug your a and b into sin(a+b).

Please be aware that this forum is made to assist you, not just to give you answers.
• March 16th 2010, 07:44 PM
pickslides
Quote:

Originally Posted by Solaris123
Please tell me the solution and the steps required to get solution to this.

Let sin(a)= 1/ sqrt(5) 0 < a < 90 degrees
cos(b)= 3/ sqrt(10) 0< b <90

then what is sin(a+b) equal to?

$\sin(a+b) = \sin(a)\cos(b)+\sin(b)\cos(a)$

And

$\sin(a)= \frac{1}{ \sqrt{5}}$

$\cos(b)= \frac{3}{ \sqrt{10}}$

To sub into the above formula.

Use $\sin^2a+\cos^2a = \sin^2b+\cos^2b =1$

To find $\sin(b)$ and $\cos(a)$
• March 16th 2010, 07:46 PM
pickslides
Quote:

Originally Posted by Anonymous1
Take the arcsine and arccosine to find a and b.
Then plug your a and b into sin(a+b).

Please be aware that this forum is made to assist you, not just to give you answers.

This exercise is supposed to make the student use trig identities to find an exact answer. Your approach while correct has no great educational purpose.
• March 16th 2010, 07:50 PM
Solaris123
I must clarify
I had forgotten to clarify that I must do this all without a calculator, making the suggestion by anonymous kinda mood. I must thanks pickles thought because now I can solve any problem that is similar, I suppose then there are similar identities for say sin(a-b) and the sort, right?
• March 16th 2010, 07:51 PM
pickslides
Quote:

Originally Posted by Solaris123
I suppose then there are similar identities for say sin(a-b) and the sort, right?

Yep, google will tell you what it is! (Rofl)