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?

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- Mar 16th 2010, 07:35 PMSolaris123[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? - Mar 16th 2010, 07:41 PMAnonymous1
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. - Mar 16th 2010, 07:44 PMpickslides

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

And

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

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

To sub into the above formula.

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

To find $\displaystyle \sin(b)$ and $\displaystyle \cos(a)$ - Mar 16th 2010, 07:46 PMpickslides
- Mar 16th 2010, 07:50 PMSolaris123I 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?

- Mar 16th 2010, 07:51 PMpickslides