Check out some similar questions!

Kendra22 · May 16, 2009, 06:54 PM

Theta and alpha are each acute angles in standard position, sin theta = 3/5 and cos alpha = 12/13

a) Determine cos (alpha+theta)
b) Determine sin (theta + alpha)

Unknown008 · May 17, 2009, 12:15 AM

Here's a list of all the basic trig identities you should know to do further trig identities:

$cosx=\frac{1}{secx}$ and $secx=\frac{1}{cosx}$

$cotx=\frac{1}{tanx}$ and $tanx=\frac{1}{cotx}$

$cosecx=\frac{1}{sinx}$ and $sinx=\frac{1}{cosecx}$

Then, the square identities;

$cos^2x + sin^2x=1$

$1 + cot^2x=cosec^2x$

$tan^2x=1+sec^2x$

Addition and subtraction of angles;

$cos(A+B)= cosAcosB - sinAsinB$ and
$cos(A-B)= cosAcosB + sinAsinB$

$sin(A+B)= sinAcosB + cosAsinB$ and
$sin(A-B)= sinAcosB - cosAsinB$

$tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$ and $tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}$

Double Angles;

$sin2A=2sinAcosA$

$cos2A = cos^2A - sin^2A$

OOOOO $=1 - 2sin^2A$

OOOOO $=2cos^2A-1$

$tan2A = \frac{2tanA}{1-tan^2A}$

Now use them to solve your questions. :)

Kendra22 · May 17, 2009, 11:56 AM

For cos (alpha + theta)
would the answer be cos (12/13 + 4/5)?

Unknown008 · May 17, 2009, 12:49 PM

No. Start by converting $cos(\alpha+\theta)$ to $cos{\alpha}cos{\theta} - sin{\alpha}sin{\theta}$

Now have a drawing, you know some of the Pythagorean triples I hope:

Then, substitute your values of sin theta, sin alpha, cos theta and cos alpha. Do the same procedure for the second one. :)

EDIT: The second triangle has a mistake.

Kendra22 · May 17, 2009, 01:03 PM

OK this is what I got
sin alpha cos theta- sinalpha sintheta
= (12/13) (4/5) - (11/13) (3/5)
= 3/13

and for the second one

sin theta cos alpha - costheta sin alpha
= (3/5)(12/13) - (4 /5 ) (11/13)
= -11/65

Unknown008 · May 18, 2009, 07:46 AM

Sorry, yesterday, it was very late and I was very tired. The second triangle has sides 12, 13, 5 instead of 12, 13, 11.

Ok you did the question well, except that you haven't notice my mistake over there. You understood the concept though, good! :) now, just re-work up the actual answers.

$sin\alpha = \frac{5}{13}$