c3 Trig help

Hi
i came across a 2 mark question in a c3 past paper

solve
arcsinX=arccosX

Not exactly sure how to do this - bearing in mind its a short 2 mark question
(it is INCORRECT to assume sinx=cosx)

Thanks

Reply 1

ombtom

17

Original post by starwarsjedi123

Hi
i came across a 2 mark question in a c3 past paper

solve
arcsinX=arccosX

Not exactly sure how to do this - bearing in mind its a short 2 mark question
(it is INCORRECT to assume sinx=cosx)

Thanks

The point where sin(x) = cos(x) is when x = 45.

So

X = 1/\sqrt2

.

Reply 2

starwarsjedi123

OP

4

Original post by ombtom

The point where sin(x) = cos(x) is when x = 45.

So

X = 1/\sqrt2

.

the examiner report says if you get 1/rt2 by stating sinx=cosx is incorrect and gets 0 marks

Reply 3

ombtom

17

Original post by starwarsjedi123

the examiner report says if you get 1/rt2 by stating sinx=cosx is incorrect and gets 0 marks

Well that's rude.

Reply 4

starwarsjedi123

OP

4

it says the correct answers substituted y in for arccosx.Not sure how that works :s-smilie:

Reply 5

Zacken

22

Original post by starwarsjedi123

it says the correct answers substituted y in for arccosx.Not sure how that works :s-smilie:

If you let

y = \arccos x

then:

Unparseable latex formula:

\displaystyle [br]\begin{align*}\cos y = x &\Rightarrow \cos^2 y = x^2 \\ & \Rightarrow 1 - \cos^2 y = 1 - x^2 \\ & \Rightarrow \sin^2 y = 1 - x^2 \\ & \Rightarrow \sin y = \sqrt{1-x^2}\end{align*}

.

So you have

\arcsin x = \arccos x \Rightarrow \arcsin x = y \Rightarrow x = \sin y \Rightarrow x = \sqrt{1-x^2} \Rightarrow \cdots

(edited 7 years ago)

Reply 6

Math12345

14

arc Cos(x) = arc Sin(x)

Let y = arc Cos(x). Then Cos(y) = x. Now x = Cos(y) = Sin(pi/2 - y). So arc Sin(x) = pi/2 - y

So we have y = pi/2 -y i.e. 2y=pi/2 so y=pi/4. Solving gives x= 1/√2.

Reply 7

C-rated

2

let sinx = cos x then divide both sides by cosx

so tan x = 1

arctan1 = x

therefore x = 45 or 225

Reply 8

ombtom

17

Original post by starwarsjedi123

it says the correct answers substituted y in for arccosx.Not sure how that works :s-smilie:

so arcsinx = y
x = siny
siny = x/1.

Draw a right-angled triangle with 1 as the hypotenuse, and x as the side opposite angle y. The third side will then be the square root of (1-x^2).

So cosy = cos(arccosx) = x.
cosy = square root of (1-x^2).

So x^2 = 1 - x^2.

Rearrange for x = 1/root2.

Reply 9

starwarsjedi123

OP

4

Original post by Zacken

If you let

y = \arccos x

then:

Unparseable latex formula:

\displaystyle [br]\begin{align*}\cos y = x &\Rightarrow \cos^2 y = x^2 \\ & \Rightarrow 1 - \cos^2 y = 1 - x^2 \\ & \Rightarrow \sin^2 y = 1 - x^2 \\ & \Rightarrow \sin y = \sqrt{1-x^2}\end{align*}

.

So you have