Trigonometry: The Inverse Trig Functions.

This post is part of a series of posts on Trigonometry. To see all the posts, click on the tag #TrigonometryTutorials below. This is the post you should read before you read this.

Now we already know the Trig functions: Sine, Cosine, Tangent, Cosecant, Secant, and Cotangent. We can find the length of a side of a Right triangle if we know an angle and one of the sides. What if we know 2 of the sides and are looking for the other components? For example, we know that the Hypotenuse is 10 and the adjacent is 6, what is the value of the angle?

This requires another type of functions: the arc functions (also known as the Inverse Trigonometric Functions). This is how the Arc Functions are defined:

sinx=y    arcsiny=x\sin x= y \iff \arcsin y = x cosx=y    arccosy=x\cos x= y \iff \arccos y = x tanx=y    arctany=x\tan x= y \iff \arctan y = x cscx=y    arccsc y=x\csc x = y \iff \text{arccsc } y = x secx=y    arcsec y=x\sec x = y \iff \text{arcsec } y = x cotx=y    arccot y=x\cot x = y \iff \text{arccot } y = x

The arc functions can also be written as: sin1\sin ^{-1}, cos1\cos ^{-1}, tan1\tan ^{-1}, csc1\csc ^{-1}, sec1\sec ^{-1}, cot1\cot ^{-1}

Back to the problem posed in the beginning: If we know that the Hypotenuse is 10 and the adjacent is 6, what is the value of the angle?

The ratios that associate the Adjacent and the Hypotenuse are Cosine and Secant:

cos=AdjHyp\cos = \frac{Adj}{Hyp} and sec=HypAdj\sec = \frac{Hyp}{Adj}.

Lets use the trig ratio Cosine because there isn't an arcsecant function on an average calculator and we want to be able to see our result. Let the angle be xx, we can now say:

cosx=610\cos x = \frac{6}{10}

Using the definition of arccosine above:

cos1610=x\cos^{-1} \frac{6}{10} = x

We can now plug in cos1610\cos^{-1} \frac{6}{10} on the calculator: we get approximately 53.13 degrees. If you want you can try the same thing with the secant function if you have the arc-secant function on your calculator.

The next post in this series is here

Note by Yan Yau Cheng
7 years, 4 months ago

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I hope you will explain in another post why your definitions of arcfunctions aren't completely correct.

Bob Krueger - 7 years, 4 months ago

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Yeah, I will do it after i explain the unit circle because i haven't introduced the trig function value of angles larger than the right angle yet

Yan Yau Cheng - 7 years, 4 months ago

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