Remaining Trigonometric ratios

[mrsrowton]

Missed test questions
#1 find the remaining trigonometric ratios of theta, given:
sec theta = sqrt 106 / 5

 

[Deleted member 4993]

Missed test questions
#1 find the remaining trigonometric ratios of theta, given:
sec theta = sqrt 106 / 5

\(\displaystyle cos(\theta) \ = \ \frac{1}{sec(\theta)}\)

\(\displaystyle sin(\theta) \ = \ \pm\sqrt{1 - [cos(\theta)]^2}\)

\(\displaystyle tan(\theta) \ = \ \pm\sqrt{[sec(\theta)]^2 \ - \ 1}\)

What else.....

 

[wjm11]

sec theta = sqrt 106 / 5

Here is another approach.

These are not hard if you draw a picture first. Draw a right triangle and label one of the acute angles as theta.

Now, we know that secant is the inverse of cosine, and cosine theta equals adjacent over hypotenuse. Therefore, secant is hypotenuse over adjacent side. Right?

We are given that “sec theta = sqrt 106 / 5”, so now we can label our triangle; make the hypotenuse “sqrt 106” and the adjacent side “5”.

We can get the third side (the opposite side) by using the Pythagorean theorem.

Once the three sides of our triangle are labeled, we can write any trig ratio asked for.