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Write the first trigonometric function in terms of the second theta for in the given quadrant:

  1. $cot\theta$
  2. $sin\theta$
  3. Where $\theta$ in Quadrant II

This problem aims to familiarize us with trigonometric functions. The concepts required to solve this problem are related to trigonometry, which includes quadrantal angles and signs of function.

The sign of a trigonometric function such as $sin\theta$ relies on the signs of the x,y coordinate points of the angle. We can also figure out the signs of all the trigonometric functions by understanding in which quadrant the angle lies. The terminal angle might lie in any of the eight regions, 4 of which are the quadrants and along the 4 axis. Each position represents something additional for the signs of the trigonometric functions.

To understand the signs of the trigonometric functions, we must understand the sign of $x$ and $y$ coordinates. For this, we know that distance between any point and origin is forever positive, but $x$ and $y$ can be positive or negative.

Expert Answer

Let’s first see the quadrants, in the $1^{st}$ quadrant, $x$ and $y$ are all positive, and all $6$ trigonometric functions will have positive values. In the $2^{nd}$ quadrant, only $sin\theta$ and $cosec\theta$ are positive. In the $3^{rd}$ quadrant, only $tan\theta$ and $cot\theta$ are positive. Ultimately, in the $4^{th}$ quadrant, only $cos\theta$, and $sec\theta$ are positive.

Now let’s start our solution since $cot\theta$ is the reciprocal of $tan\theta$, which is equal to $\dfrac{$sin\theta$}{ $cos\theta$}$, so:

\[cot\theta = \dfrac{cos\theta}{sin\theta}\]

To rewrite $cot\theta$ only in terms of $sin\theta$, we have to change $cos\theta$ into $sin\theta$, using the trigonometric identity:

\[cos^2 \theta + sin^2 \theta = 1\]

\[cos^2 \theta = 1 – sin^2 \theta\]

\[cos\theta = \pm \sqrt{1 – sin^2 \theta}\]

Since $cos\theta$ lies in the $2^{nd}$ quadrant, we will apply the negative sign to equal its effect:

\[cot\theta = \dfrac{-cos\theta}{sin\theta}\]

\[cot\theta = \dfrac{- \sqrt{1 – sin^2 \theta}}{sin\theta}\]

Hence, this our final expression of $cot\theta$ in terms of $sin\theta$.

Numerical Result

The final expression of $cot\theta$ in terms of $sin\theta$ is $\dfrac{- \sqrt{1 – sin^2 \theta} }{sin\theta}$.

Example

Write $tan\theta$ in terms of the $cos\theta$, where $\theta$ lies in the $4$ Quadrant. Also write other trigonometric values in Quad III for $sec\theta = -2$.

Part a:

Since $tan\theta$ is the fraction of $sin\theta$ over $cos\theta$, so:

\[tan\theta=\dfrac{sin\theta}{cos\theta}\]

To write in terms of $cos\theta$, applying the change using the trigonomteric identity:

\[cos^2 \theta + sin^2 \theta = 1 \]

\[sin^2 \theta = 1 – cos^2 \theta \]

\[sin\theta = \pm \sqrt{1 – cos^2 \theta} \]

Since $sin\theta$ lies in the $4^{th}$ quadrant, apply negative sign :

\[tan\theta = \dfrac{-sin\theta}{cos\theta} \]

\[tan\theta = \dfrac{-\sqrt{1 – cos^2 \theta}}{cos\theta} \]

Part b:

Using the definition of $secant$:

\[sec\theta = \dfrac{hypotenuse}{base}\]

To find the other sides of the right triangle we will use the Pythagorean theorem:

\[H^2 = B^2 + P^2 \]

\[P = \sqrt{B^2 – H^2}\]

Since $sec$ lies in the III Quad, we will apply the negative sign:

\[ P = -\sqrt{2^2 + 1^2}\]

\[ P = -\sqrt{3}\]

Now find the other values:

\[ sin\theta = -\dfrac{\sqrt{3}}{2}\]

\[ cos\theta = -\dfrac{1}{2}\]

\[ tan\theta = \sqrt{3}\]

\[ cot\theta = \dfrac{\sqrt{3}}{3}\]

\[ cosc\theta = -\dfrac{2\sqrt{3}}{3}\]

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