Expert: Scotto - 6/20/2009

Question
Hi,
My name's Veronica and I'm currently having trouble
answering these problems. I believe they have to be verified
using the pythagorean identities (i think) but my notes are
only about the sin, cos, tan, etc. I dont have notes about
these kind of problems. Please help me. Thanks.

Problems:

1) Verify: sin(theta)cos(theta)=x(square root of 25-x^2) all
over 25. where theta is an acute angle of triangle ABC withe
one side x.

2) tan^2A=x^2 all over 9-x^2

3) 4sin^2A over cosA= x^2 all over square root of 16-x^2

I will Be waiting for your answers.. Please help me I would
greatly appreciate it.. Thanks again.

Answer
1) Verify: sinΘcosΘ = x√(25 - x²)/25.
Where theta is an acute angle of triangle ABC with one side x.

It is known that for the right triangle with legs a=√(25-x²) and b=x,
the hypoteneuse is is c and a² + b² = c².  Note also that the value I gave of a and b could be interchanged and the results would still be the same.

Given this, c = √(25 - x² + x²) = √25 = 5.

It is also known that sinΘ = b/c abd cosΘ = a/c.

We know that b/c = x/5 and that a/c = √(25 - x²)/5.

If they are multiplied together, the answer is x√(25 - x²)/5²,
which is the same as x√(25 - x²)/25.


2) tan^2A=x^2 all over 9-x^2

Let a=x and c=3.  Given that a² + b² = c²,

we know that b=√(3²-x²)=√(9-x²).

This is because a² + b² = c² is true since a²=x² and b²=9-x².

As can be seen, when these are added together, all that is left is 9.

It can also be seen that 9=3².


3) 4sin^2A/cosA= x^2 all over square root of 16-x^2

Take a=x and c=4.  Since a² + b² = c², we know that b = √(c² - a²).

With the variable a being the far side, we can say that 4sin²A = 4(a²/c²) = 4(x²/4²) = x²/4.

We also can say that cosA = √(16-x²) / 4.

Combining both of these, it can be said that
4sin²A/cosA = (x²/4) / (√(16-x²) / 4).

Since both the numerator and denominator contain /4, they cancel.

This leaves x² / √(16-x²).