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Given tanj=5/12 find cosg

User Kitfox
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Given tan g = 5/12, find cos g:

5
tan g = ———
12

The tangent function is, by definition, the quotient between sine and cosine:

sin g 5
———— = ———
cos g 12

Product of them extremes = product of the means

12 · sin g = 5 · cos g

Square both sides:

(12 · sin g)² = (5 · cos g)²

12² · sin² g = 5² · cos² g

144 · sin² g = 25 · cos² g

But sin² g = 1 – cos² g. Substitute it for sin² g into the equation above, and you have

144 · (1 – cos² g) = 25 · cos² g

Multiply out the brackets, and then isolate cos² g:

144 – 144 · cos² g = 25 · cos² g

144 = 25 · cos² g + 144 · cos² g

144 = 169 · cos² g

Divide both sides by 169:

144
cos² g = ———
169

12²
cos² g = ———
13²

cos² g = (12/13)²

Now, take the square root of both sides:

cos g = ± √(12/13)²

12
cos g = ± ———
13

The sign of cos g depends on which quadrant the angle g lies. As tan g = 5/12, which is positive, then g lies either in the 1st or the 3rd quadrant:

• If g lies in the 1st quadrant, then

5
cos g > 0 ⇒ cos g = ——— ✔
13

• If g lies in the 3rd quadrant, then

5
cos g < 0 ⇒ cos g = – ———
13


I hope this helps. =)


Tags: trigonometric relation tangent cosine sine tan cos sin trig trigonometry

User Suamikim
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