You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Skylor****Guest**

given cos θ = 40/41, θ lies in quadrant IV

find sin 2θ, cos 2θ, and tan 2θ

**zetafunc.****Guest**

Can you spot the Pythagorean triple here?

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,613

hi Skylor,

From the look of this question, you are expected to do this without a calculator.

Set adjacent = 40 and hypotenuse = 41 . Then work out the 'opposite' using Pythagoras.

(As the angle is not acute you may wonder if this is valid but, at this stage you may treat it as if it is acute)

That means you can write down a value for sin θ.

But now you must adjust your answer (+ or -) according to which quadrant this angle lies in.

After that you can get the angles you want with

sin2θ = 2sinθcosθ

cos2θ= (cosθ)^2 - (sinθ)^2

tan 2θ = 2tanθ/( 1 - [tanθ]^2 ) or sin2θ/cos2θ

Of course you can also check your answers using a calculator.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**skylor****Guest**

This is for a precalc class. We are allowed to use calculators. I know this is going to be on a quiz so I want to make sure I'm doing it right.

Pages: **1**