Pre-calc, PLEASE HELP

[evolchild9]

okay I just wanted to see if these problems i did were correct, so if anyone could tell me if they are, and also if they are wrong, how to do them correctly?

Solve for theta values in Q1.

1.) sin^2 theta=25/26, find cos theta.

I got the answer cos=1/squareroot{26}.

2.) tan^2 theta=1/4, find sec^2 theta.

I got the answer sec^2 theta=squareroot{17}/4


Express each value of the angle of the function in Q1.

1.) sin(600 degrees)

I got -sin60

2.) cos(-562 degrees)

I got -cos22

3.) tan(-350 degrees)

I got tan10

4.) csc(850 degrees)

I got csc50


Simplify as much as possible.

csc theta/1+cot^2 theta

I got sin theta

 

[soroban]

Hello, evolchild9!

Great work . . . nice going!

Just one slip-up on #2. . . . well, two.

2.) tan²θ = 1/4, find sec²θ.

I got the answer: sec² theta = sqrt{17}/4
.

You forgot the <u>square</u> on the tangent: . tan θ = 1/2

Even if your work was correct, you forgot to <u>square</u> the secant.


The rest was excellent!

 

[evolchild9]

I'm a tad confused at what you mean i should do, in terms of squaring. I figured that since the tan was squared and they asked for the secant to be squared in the answer that i could just find the answer the same way as normally through sohcatoh, and the triangle.

 

[Gene]

I suspect they wanted you to use the identity
sec²(theta) = 1 + tan²(theta)
You can at least use it to check your answer if you want to do it some other way. It isn't hard to prove.

 

[evolchild9]

I feel really dumb, but i tried that just now, but i still get the same answer with sec^2 theta =Squareroot{17}/4

1+ (1/4)^2=sec^2 theta
1+1/16=sec^2 theta
16/16+1/16=sec^2 theta
17/16=sec^2 theta
squareroot {17}/squareroot{16}=sec^2 theta
Squareroot{17}/4=sec^2 theta

 

[Gene]

The 1/4 is already squared. It is tan squared so the answer is
1+1/4 = 5/4

 

[evolchild9]

Thanks so much