# trigo > i cant do it!

• Sep 9th 2008, 05:03 AM
sanikui
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!
• Sep 9th 2008, 05:23 AM
mr fantastic
Quote:

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).
• Sep 9th 2008, 05:33 AM
sanikui
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
• Sep 9th 2008, 05:43 AM
mr fantastic
Quote:

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.