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## Need help with 3 Fundamental Inverse Trig Questions

I was told these 3 questions were crucial to the understanding of the topic, but I can't manage to work them out

(a) Find the exact value of sin-1(sin 5&pi;/4)
(b) Why is the answer -&pi;/4 and not 5&pi;/4?
(a) Find sin(sin-1(-root2))
(b) Why is the answer undefined?
(a) Find tan(2(sin-1 5/13))

Question 1

The key to question 1 is that inverse trigonometric functions generally only have one cycle. Unlike trigonometric functions which have a period where that one cycle is unique, followed by an infinite series of repeated cycles, an inverse trigonometric function only exists for one cycle.

Note that most texts will refer to the inverse trigonometric function with one cycle with capital letters.

(i.e.: Sin-1, Cos-1 and Tan-1, as opposed to sin-1, cos-1 and tan-1)

It's clear from the questions that they meant the one-cycle inverse functions though (otherwise they wouldn't be 'functions' &ndash; one y-value for any x-value &ndash; in the VCE sense of the word)

The cycle you choose is arbitrary, but for the sake of convention:

Sin-1 runs from -&pi;/2 to &pi;/2 (range)
Cos-1 runs from 0 to &pi; (range)
Tan-1 runs from -&pi;/2 to &pi;/2 (range)

Hence why the answer to part (a) is -&pi;/4 instead of 5&pi;/4. (You're converting it into a Q4 angle, and then subtracting 2&pi; to make it range from -&pi;/2 and 0)

Question 2

The sine inverse function is asking "what value of sine would give me -root2?" Since -root2 roughly equals -1.4142, which is less than -1, there is no such value of sine that can do this!

Question 3

The key to question 3 is to treat inverse trigonometric functions as angles. After all, you input the result of a trigonometric function, and it spits out the angle (does the reverse of a trigonometric function).

Start by letting &alpha; = sin-1(5/13) which gives sin(&alpha;) = 5/13

Using SOHCAHTOA:

This means opposite over hypotenuse is 5/13. Let o = 5, h = 13.
Therefore, a = 12
So tan(&alpha;) = 5/12

Using the double angle formula: tan(2&alpha;) = 2tan(&alpha;) / (1 - tan2(&alpha;))

tan(2&alpha;) = 2(5/12) / (1 - 25/144) = 120/119

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