Using the identity tan⁡θ=tan⁡(θ−180∘) to find values of θ such that tan⁡θ=tan⁡20∘

Since tan⁡θ=tan⁡(nπ+θ), where n is any integer and θ=π9 in our case. So, we have 0≤nπ+π9≤2π⇒0≤(9n+1)π≤18π. Now you can clearly see that only n=0 and n=1 satisfy this bound, hence the only answers are θ=π9,10π9

Hint:period of tangent function is 180∘,tan⁡20∘=tan⁡(180+20)∘,tan⁡(20−180)∘=tan⁡(−160)∘,200∘ is anticlockwise rotation, −160∘ is clockwise rotation.

θ=20∘;θ=20∘±n⋅180∘; Other angles are =20∘,200∘,380∘,460∘ We can add or subtract n⋅π ( n positive or negative integer) to 20∘ getting the tip of radius vector from first to third quadrant. The radius vector can be rotated indefinitely around the origin adding 180∘

Since tan⁡θ=tan⁡(nπ+θ), where n is any integer and θ=π9 in our case. So, we have 0≤nπ+π9≤2π⇒0≤(9n+1)π≤18π. Now you can clearly see that only n=0 and n=1 satisfy this bound, hence the only answers are θ=π9,10π9

Hint:period of tangent function is 180∘,tan⁡20∘=tan⁡(180+20)∘,tan⁡(20−180)∘=tan⁡(−160)∘,200∘ is anticlockwise rotation, −160∘ is clockwise rotation.

θ=20∘;θ=20∘±n⋅180∘; Other angles are =20∘,200∘,380∘,460∘ We can add or subtract n⋅π ( n positive or negative integer) to 20∘ getting the tip of radius vector from first to third quadrant. The radius vector can be rotated indefinitely around the origin adding 180∘

"Using the identity tan⁡θ=tan⁡(θ−180∘) to find values of..." was answered by our experts

Joan Thompson

Answered question

2022-01-15

Using the identity

\tan θ = \tan (θ - 180^{\circ})

to find values of

θ

such that

\tan θ = {\tan 20}^{\circ}

Answer & Explanation

Annie Levasseur

Beginner2022-01-18Added 30 answers

Since

\tan θ = \tan (n π + θ)

, where n is any integer and

θ = \frac{π}{9}

in our case. So, we have

0 \leq n π + \frac{π}{9} \leq 2 π \Rightarrow 0 \leq (9 n + 1) π \leq 18 π

. Now you can clearly see that only n=0 and n=1 satisfy this bound, hence the only answers are

θ = \frac{π}{9}, \frac{10 π}{9}

RizerMix

Expert2022-01-20Added 656 answers

Hint:period of tangent function is

180^{\circ}, \tan 20^{\circ} = \tan (180 + 20)^{\circ}, \tan (20 - 180)^{\circ} = \tan (- 160)^{\circ}, 200^{\circ}

is anticlockwise rotation,

- 160^{\circ}

is clockwise rotation.

user_27qwe

Skilled2022-01-23Added 375 answers

θ = 20^{\circ}; θ = 20^{\circ} \pm n \cdot 180^{\circ}

; Other angles are

= 20^{\circ}, 200^{\circ}, 380^{\circ}, 460^{\circ}

We can add or subtract

n \cdot π

( n positive or negative integer) to

20^{\circ}

getting the tip of radius vector from first to third quadrant. The radius vector can be rotated indefinitely around the origin adding

180^{\circ}

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