# Thread: trigo > i cant do it!

1. ## trigo > i cant do it!

Find the required expression without using a calculator :
1. sin2A, give sinA=3/5
2. cos2A, given cosA=3/5
3. tan2A, given tanA=1/2

Now my problem is use the trigo identities to find sin2A. cos2A and tan2A. i know can use triangle to find, but problem is not allow use triangle to find exact value.

THX who Helping!

2. Originally Posted by sanikui
Find the required expression without using a calculator :
1. sin2A, give sinA=3/5
2. cos2A, given cosA=3/5
3. tan2A, given tanA=1/2

Now my problem is use the trigo identities to find sin2A. cos2A and tan2A. i know can use triangle to find, but problem is not allow use triangle to find exact value.

THX who Helping!
Are you familiar with the double angle formulae?

1. sin A = 3/5 => cos A = 4/5 (assuming 1st quadrant).

sin (2A) = 2 sin A cos A = ....

2. Done in a similar way to 1. Use the formula for cos (2A).

3. Substitute tan A = 1/2 into the formula for tan (2A).

3. Like this question...
Find, without the use of a calculator, the value of sinA, given that cos2A=5/13

Solution:
cos(2A)=1-2sin^(2A)
2sin^(2A)=1-cos(2A)
sin^(2A)=1/2(1-cos(2A)) =1/2(1-5/13)
sin(A)=2 square root 13/13

4. Originally Posted by sanikui
Like this question...
Find, without the use of a calculator, the value of sinA, given that cos2A=5/13

Solution:
cos2A=1-2sin^2A
2sin^2A=1-cos2A
sin^2A=1/2(1-cos2A) =1/2(1-5/13)
sinA=2 square root 13/13
So what's the problem? The only thing I'd add is a justification for why the negative solution for sin A is rejected.