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# c3 Trig help Watch

1. 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
2. (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 .
3. (Original post by ombtom)
The point where sin(x) = cos(x) is when x = 45.

So .
the examiner report says if you get 1/rt2 by stating sinx=cosx is incorrect and gets 0 marks
4. (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.
5. it says the correct answers substituted y in for arccosx.Not sure how that works
6. (Original post by starwarsjedi123)
it says the correct answers substituted y in for arccosx.Not sure how that works
If you let then:

.

So you have
7. 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.
8. let sinx = cos x then divide both sides by cosx

so tan x = 1

arctan1 = x

therefore x = 45 or 225
9. (Original post by starwarsjedi123)
it says the correct answers substituted y in for arccosx.Not sure how that works
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.
10. (Original post by Zacken)
If you let then:

.

So you have
it wont let me rep you ..but *REP*
11. (Original post by C-rated)
let sinx = cos x then divide both sides by cosx

so tan x = 1

arctan1 = x

therefore x = 45 or 225
yeah m8 thats incorrect

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