C3 Trig - Help please!

How do you do the question I above? I assume you put it equal to pi/3 and -pi/3, without the modulus? I just don't know how to solve it from there on.

Reply 1

CalculusMan

You have arcsin(x - 1) = +- Pi/3

Take sin's of each side.

Reply 2

Coda

Original post by CalculusMan

You have arcsin(x - 1) = +- Pi/3

Take sin's of each side.

That's where I got stuck. How do I "take sins" exactly? Does it become " x-1 = sin(pi/3) "?

Reply 3

M1K3

The equations in blue you have written are correct :smile:

, lets label the top one A, and the bottom one B.

Lets multiply both equations by

-1

Coming to;

A \Rightarrow sin^{-1}(x-1) = -\frac{\pi}{3}

B \Rightarrow sin^{-1}(x-1) = \frac{\pi}{3}

Working at A, get the sin to the other side,

A \Rightarrow x-1 = sin(-\frac{\pi}{3})

which solves to

A \Rightarrow x = 1 + sin(-\frac{\pi}{3})=1- \frac{\sqrt 3}{2}

do the same for the other equation which would get the same except opposite sign.

Reply 4

CalculusMan

Yes, and similarly for the other equation.

sin(arcsin(x)) = x for x between -1 and 1.

Reply 5

M1K3

Original post by Coda

That's where I got stuck. How do I "take sins" exactly? Does it become " x-1 = sin(pi/3) "?

Yes thats true think of it like,

y = sin x

therefore

sin^{-1} y = x

Or, think of it like this, to calculate the sin of 30 degrees you would due

sin 30 = 0.5

When you have 0.5 and want to find the angle, on your calculator you would use,

sin^{-1} 0.5 = 30

Its more like the opposite action using the inverse.

Quick Reply

Latest

Articles for you

What kind of university is best for you?

How our community record system works

The Student Room community guidelines

Adding your personal statement to the library

C3 Trig - Help please!

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you