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Can anybody proof

"the rules used to find trigonometric functions of angles greater than $90^{\circ}$"

i.e. if angle$= ( nπ ± θ)$, then its trigonometric function $=$ trigonometric function of $θ$ and sign is decided by ASTC (CAST) rule ?

nmasanta
  • 9,222
Tushar
  • 147

2 Answers2

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Let $\theta$ be in the first quadrant, so sin$(\theta)$ and cos$(\theta)$ are positive.

Then tan($\theta)=\dfrac{\sin(\theta)}{\cos(\theta)}$ is positive too.

Now $\theta+\dfrac\pi2$ is in the second quadrant. Using the trigonometric addition formulas,

$\sin\left(\theta+\dfrac\pi2\right)=\sin(\theta)\cos\left(\dfrac\pi2\right)+\cos(\theta)\sin\left(\dfrac\pi2\right)=\cos(\theta) $ is positive,

$\cos(\theta+\dfrac\pi2)=\cos(\theta)\cos(\dfrac\pi2)-\sin(\theta)\sin(\dfrac\pi2)=-\sin(\theta)$ is negative,

and therefore tan$\left(\theta+\dfrac\pi2\right)=\dfrac{\sin\left(\theta+\dfrac\pi2\right)}{\cos\left(\theta+\dfrac\pi2\right)}$ is negative in the second quadrant.

Similar arguments work for $\theta+\pi$ in the third quadrant:

$\sin(\theta+\pi)=-\sin(\theta)$ and $\cos(\theta+\pi)=-\cos(\theta)$ are negative, so their ratio $\tan(\theta+\pi)$ is positive.

I'm leaving the fourth quadrant $\left(\theta+\dfrac{3\pi}2\right)$ as an exercise for the reader.

J. W. Tanner
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From a logical standpoint (not proof as such), I understood it this way:

1st quadrant unit circle, x is positive, y is positive and hence all trigonometric functions are positive

2nd quadrant unit circle, x is negative, y is positive
sin = y / 1 = y
cos = - x / 1 = -x
tan, because one of the sin and cos is negative tan will be negative

3rd quadrant unit circle,
x is negative, y is negative
sin = -y / 1 = - y
cos = -x / 1 = -x
since tan is a division of sin and cos, negative cancels out and is positive.

4th quadrant unit circle,
x is positive, y is negative
sin = -y / 1 = -y
cos = x / 1 = x
since one of the sin and cos is negative, tan is negative

Sumedh Rao
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