Is this mathematically correct?

[Ricochet J]

Suppose sinΘ=0.6

Then Θ=arcsin0.6

So, I should be able to say cos(arcsin0.6) to get my answer in terms of cos

I should get 0.8 but I keep getting it wrong! I keep getting an unknown value with many decimal places on my claculator

 

[Tommy B]

We got this in a C4 practice paper and you can't just do Cos(arcsin0.6)

I can't remember the method. Hopefully someone will post it as it'll be useful revision for me

 

[qwerty]

Suppose sinΘ=0.6

Then Θ=arcsin0.6

So, I should be able to say cos(arcsin0.6) to get my answer in terms of cos

I should get 0.8 but I keep getting it wrong! I keep getting an unknown value with many decimal places on my claculator

Make sure your calculator is set to radians and not degrees.

 

[Ricochet J]

^^ Actually, I was in radians and I wanted it degrees. When I do it in degrees I get 0.8!

Damn radians

 

[daz]

Degrees are for the lose.

 

[//Mike]

Question is find cos(t)? Much easier to consider a right angled triangle with the hypotenuse as length 1, opposite side length of 0.6 - leaving you with an unknown side which is equal to cos(t)...

 

[hendrix]

Works the same in Degrees Are you putting brackets around the asin0.6 ? Depending on how your calculate handles cosines/sines, just gotta make sure you keep the (asin0.6) inside brackets and you should get 0.8.

 

[Psiko]

Either look at a right-angled triangle. or use the identity cos^2 t + sin^2 t = 1.

Consider a 3-4-5 triangle, where the angle, t, being considered is between the sides of length 4 and 5. The sint = 3/5 and cost = 4/5.

 

[Duff-Man]

An easier way to interchange cos and sin is to use:

cos^2(x) + sin^2(x) = 1

hence: cos(x) = sqrt(1 - sin^2(x)) etc

 

[Dave]

This is C4? I thought this was elementary trig. (it was in IB maths . . .)

I'd use the "triangle method" of constructing a triangle and using the given angle to put on 2 sides, and then use cos = Adj/Hyp as mentioned above.