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The Outer Function

Part of the Trig Equations and Inverses section of Coddy's Trigonometry journey. Lesson 12 of 52.

Find sin⁡(arccos⁡(35))\sin\left(\arccos\left(\frac{3}{5}\right)\right). The inner angle has cos⁡(θ)=35\cos\left(\theta\right) = \frac{3}{5}: adjacent 3, hypotenuse 5. The missing side is the opposite one, and Pythagoras makes it 4.

Read the outer function off the triangle: sine is opposite over hypotenuse, so sin⁡(arccos⁡(35))=45\sin\left(\arccos\left(\frac{3}{5}\right)\right) = \frac{4}{5}. Which side was missing depends on the inner function; which ratio you read depends on the outer one. Keep the two jobs apart.

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Challenge

Find the exact value of y=tan⁡(arccos⁡(45))y = \tan\left(\arccos\left(\frac{4}{5}\right)\right). Draw the triangle on paper, then write y = … as a fraction and press Check.

Try it yourself

y=tan⁡(arccos⁡(45))y = \tan\left(\arccos\left(\frac{4}{5}\right)\right)

Get yy alone

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