marcusay
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Hi all!

I'm teaching myself A-Level maths and my online tutor is just... no help. Like most, I'm finding trig to be the biggest curveball in terms of comprehension so far and I struggle answering a lot of the basics. Could I get some help with the following question please?

If α is an acute angle, find three possible values in terms of α for θ, where –360° ≤ θ ≤ 360° and

(a) sin θ = sin α
(b) tan θ = tan α
(c) cos θ = –cos α
(d) sin θ = sin (180° + α)
(e) tan θ = tan (180° – α)
(f) cos θ = cos (360° – α)

I know beggars can't be choosers, but ideally a step-by-step breakdown of what to do would be grand so I'd be able to develop a technique for myself. Thanks everyone!
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the bear
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so i would choose an angle for α, let us say 30°.... so sinα = 0.5, tanα = 1/√3, cosα = (√3)/2

i) solve the equation sinθ = 0.5

ii) solve the equation tanθ = 1/√3

iii) solve the equation cosθ = - (√3)/2

etc
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marcusay
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(Original post by the bear)
so i would choose an angle for α, let us say 30°.... so sinα = 0.5, tanα = 1/√3, cosα = (√3)/2

i) solve the equation sinθ = 0.5

ii) solve the equation tanθ = 1/√3

iii) solve the equation cosθ = - (√3)/2

etc
Thanks for the help! But, I need to give my answers in terms of α for θ (i.e. θ = 180 + α or whatever) and for the whole range of –360° ≤ θ ≤ 360°.
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RDKGames
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(Original post by marcusay)
Hi all!

I'm teaching myself A-Level maths and my online tutor is just... no help. Like most, I'm finding trig to be the biggest curveball in terms of comprehension so far and I struggle answering a lot of the basics. Could I get some help with the following question please?

If α is an acute angle, find three possible values in terms of α for θ, where –360° ≤ θ ≤ 360° and

(a) sin θ = sin α
(b) tan θ = tan α
(c) cos θ = –cos α
(d) sin θ = sin (180° + α)
(e) tan θ = tan (180° – α)
(f) cos θ = cos (360° – α)

I know beggars can't be choosers, but ideally a step-by-step breakdown of what to do would be grand so I'd be able to develop a technique for myself. Thanks everyone!
You need to know basic properties like periodicity and identities for each trig function.

For instance, we are told that \alpha is acute, this means that 0 < \alpha < 90. One obvious solution to the first question is \boxed{\theta = \alpha}.
However, you should also be aware of the identity \sin \alpha \equiv \sin(180-\alpha). Therefore \boxed{\theta = 180-\alpha} is another solution. It's important for you to note that -360 \leq \theta \leq 360 must hold, and so for these solutions, you should understand how indeed they lie in this interval. Any solution not lying entirely inside this interval would be rejected.

The third solution comes from knowing that sine repeats every 360 degrees. This means that \sin \alpha \equiv \sin (\alpha - 360). What's the final solution for \theta in terms of \alpha given this fact?? Does it lie entirely in our required interval?
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