SP# 7: Unit Q Concept 2: Find all Trig Functions when given one trig function and quadrant (using identities and SOH CAH TOA)

Finding All Trig Functions (with identities and SOH CAH TOA)

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Given secx=-rad185/8 ; tanx= >0 . Find all the trigonometric functions.

Solved with Pythagorean/Ratio/Reciprocal Identities (Unit Q Concept 2): *note: The identity work is written in red letters.

First, we need to figure out in what quadrant our triangle is in given the clues. We must know this so we get the sign of our trigonometric functions correct. Based on our clues, our triangle lies in quadrant 2 meaning only tangent and cotangent will be positive, the rest will be negative.

Given secant, it is easy to find its inverse, cosine. What we do is use the reciprocal identity of secx=1/cosx.

Using the Pythagorean identity of 1+tan^2x=sec^2x we can find tangent.

Another Pythagorean identity which states that sin^2x+cos^2x=1 allows us to find cosine. We were then left with a value that could be positive or negative but from what we did in the beginning, we know sine must be negative.

Using the reciprocal identity of cscx=1/sinx we can find cosecant because we have already found sine.

We can then use the tangent which was already found in green writing to find cotangent using the reciprocal identity of cotx=1/tanx.

Our findings correspond with the prediction that the triangle lied in quadrant 2 and the signs correspond also.

Solved with right triangles [SOH CAH TOA] (Unit O Concept 5) *note: The SOHCAHTOA work is explained along with each picture.

Knowing Secant is -rad185/8 we can find cosine because we have both hypotenuse and adjacent. Our answer is negative, which corresponds with what we found in the beginning (that our triangle is in quadrant 2) since cosine is negative there.

From here we can use Pythagorean theorem to find our missing side, opposite. We find that it could be 11. 

Then we can find sin, now that we found opposite and were given hypotenuse.

Its inverse would be cosecant and is found by setting sine over one and rationalizing.

Tangent and cotangent are easily found because they are opposite over adjacent for tangent and adjacent over opposite for cotangent and we have already found the values for adjacent and opposite using what was given and the Pythagorean theorem.

From our answers and what we concluded given the clues we can conclude that our triangle looks something like this: