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Mod edit: Don't post your age on the forum, peoples...

I am taking some gifted classes. I am preparing for my final exam, and I have found some problems that I do not understand how to do. Can somebody help me with them? Here they are:

1) tan(sin^-1(24/25))
(sin^-1 is the inverse sine)

2) (in degrees) 2cos^2(x)+3sin(x)-1=0

3) sin(x)(cos(x))=1/2

4) 3sinx=1+cos2x

5) Prove that the following equation is an identity:
sin^4(x)-cos^4(x)=-cos(2x)

Thanks.

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I just need a bit of help, not flames...

WTF is this supposed to say?

They're not very hard at all, you should be able to do them.

Try drawing out the triangles in each case, or using identities such as sin^2(x) + cos^2(x) = 1.

1) ok, sin^-1(24/25) --> sinè = 24/25

so draw the triangle…

/ |
/ |
25 / | 24
/ |
/ |
/_*____|

7

(theta is marked with the star)

sinè = opposite/ hypotenuse = 24/25
tanè = opposite/adjacent = 24/7

Hence tan(sin^-1(24/25))=24/7 = 3.42857…

2) Substitute cos^2(x) for (1- sin^2(x))…

2(1 - sin^2(x)) + 3sinx – 1 = 0
2 – 2sin^2(x) + 3sinx – 1 = 0
-2sin^2(x) + 3sinx + 1 = 0
2sin^2(x) – 3sinx – 1 = 0

sin(x) = y

2y^2 – 3y – 1 = 0

By quadratic formula,

y = 1.780776… or y = -0.280776…

Hence x = sin^-1(1.780776) = no solutions
x = sin^-1(-0.280776) = -16.53° (and 180 - - 16.53 = 196.53°)

3) 2sin(x)cos(x) = 1
sin(2x) = 1 (using double angle formula)
2x = 90
x = 45

4) Use the double angle formula to expand cos(2x) into 1-sin^2(x)

This gives:

3sin(x) = 1 + 1 – sin^2(x)
sin^2(x) + 3sin(x) = 2
sin^2(x) + 3sin(x) – 2 = 0
(sin(x) – 2)(sin(x) – 1) = 0

Hence sin(x) = 1 or 2

Sin(x) = 2 --> no solutions
Sin(x) = 1 --> x = 90° or ð radians

5) I’ll need more time to do this one, sorry!