Question

Find the exact values of the six trigonometric functions of given the point (-4, 5) on the terminal side of in standard position.

Answer 1

Answer:

$\sin(\theta)=5\sqrt{41}/41\text{ and } \csc(\theta)=\sqrt{41}/5\\\cos(\theta)=-4\sqrt{41}/41\text{ and } \sec(\theta)=-\sqrt{41}/4\\\tan(\theta)=-5/4\text{ and } \cot(\theta)=-4/5$

Step-by-step explanation:

Please refer to the attached figure.

So, we can see that our angle θ is in QII.

Recall All Students Take Calculus. Since this is QII, we use the Students. In other words, only sine (and cosecant) is positive. So, cosine and tangent are negative.

Now, we also know that a point is (-4,5). Referring to our figure, this means that our adjacent side is 4 (technically -4, but we can ignore this) and our opposite side is 5. So, to find the other ratios, let's find the hypotenuse.

Use the Pythagorean Theorem:

$a^2+b^2=c^2$

Substitute 4 for a and b for 5. Solve for c. So:

$4^2+5^2=c^2$

Square:

$16+25=c^2$

Add:

$c^2=41$

Take the square root:

$c=\sqrt{41}$

So, our side lengths are: Opposite=5; Adjacent=4; and Hypotenuse=√41.

Now, we can find our side lengths.

Sine and Cosecant:

$\sin(\theta)=opp/hyp$

Substitute 5 for Opp and √41 for Hyp. So:

$\sin(\theta)=5/\sqrt{41}$

Rationalize:

$\sin(\theta)=5\sqrt{41}/41$

Since our angle is in QII, sine stays positive.

Cosecant is the reciprocal of sine. So:

$\csc(\theta)=\sqrt{41}/5$

Cosine and Secant:

$\cos(\theta)=adj/hyp$

Substitute 4 for Adj and √41 for Hyp:

$\cos(\theta)=4/\sqrt{41}$

Rationalize:

$\cos(\theta)=4\sqrt{41}/41$

Since our angle is in QII, cosine is negative. So:

$\cos(\theta)=-4\sqrt{41}/41$

Secant is the reciprocal of cosine. So:

$\sec(\theta)=-\sqrt{41}/4$

Tangent and Cotangent:

$\tan(\theta)=opp/adj$

Substitute 5 for Opp and 4 for Adj. So:

$\tan(\theta)=5/4$

Since our angle is in QII, tangent is negative. So:

$\tan(\theta)=-5/4$

Cotangent is the reciprocal of tangent:

$\cot(\theta)=-4/5$

And we are finished!