How good are you at Trigonometry?

Geometry Level 1

Find the value of θ \theta in the range 0 ≤ θ ≤ 18 0 ∘ 0\leq \theta \leq 180^ {\circ} such that

sin ⁡ θ = cos ⁡ θ \sin \theta = \cos \theta

All angles are in Degrees.


The answer is 45.

Mehul Arora by Mehul Arora , 20, India

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1 solution

Nihar Mahajan
Jul 12, 2015

sin ⁡ θ = cos ⁡ ( 9 0 ∘ − θ ) = cos ⁡ θ ⇒ 9 0 ∘ − θ = θ ⇒ 9 0 ∘ = 2 θ ⇒ θ = 4 5 ∘ \sin\theta=\cos(90^\circ-\theta)=\cos\theta \\ \Rightarrow 90^\circ-\theta=\theta \\ \Rightarrow 90^\circ = 2\theta \\ \Rightarrow \Large\boxed{\theta=45^\circ}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Not quite complete. Why can't we have sin ⁡ ( θ ) = cos ⁡ ( 45 0 ∘ − θ ) \sin(\theta) = \cos(450^\circ - \theta) ? Or sin ⁡ ( θ ) = cos ⁡ ( 27 0 ∘ + θ ) \sin(\theta) = \cos(270^\circ + \theta) ?

Your second step should show that 9 0 ∘ − θ + n ⋅ 36 0 ∘ = θ 90^\circ - \theta + n\cdot 360^\circ = \theta for integer n n . Then you should show that only one solution exist in for the given range of θ \theta .

And of course, there's a much simpler approach to this.

Hint : Divide. And why can't we divide?

Or we can just intepret from the graphs of sine and cosine functions.

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