Right triangles and identities

The six trigonometric ratios, special angles, simplifying expressions and proving identities. 10 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution. In all answers \(k, m \in \mathbb{Z}\).

Problem 1 Six trigonometric ratios

In a right triangle \(ABC\), the right angle is at \(B\), \(AB = 5\), \(AC = 13\) and \(m(\angle ACB) = x\). Find the six trigonometric ratios of \(x\).

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By the Pythagorean theorem \(BC = \sqrt{13^2 - 5^2} = \sqrt{144} = 12\).

For the angle \(x\) at \(C\): the opposite side is \(AB = 5\), the adjacent side is \(BC = 12\), the hypotenuse is \(AC = 13\).

\[\sin x = \frac{5}{13},\quad \cos x = \frac{12}{13},\quad \tan x = \frac{5}{12},\]

\[\cot x = \frac{12}{5},\quad \sec x = \frac{13}{12},\quad \csc x = \frac{13}{5}.\]

Problem 2 Ratios from one ratio

In a right triangle, \(\sin x = \frac{3}{5}\). Find \(\dfrac{\cos x + \tan x}{\cot x}\).

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Take the opposite side 3 and the hypotenuse 5. Then the adjacent side is \(\sqrt{25 - 9} = 4\).

\(\cos x = \frac{4}{5}\), \(\tan x = \frac{3}{4}\), \(\cot x = \frac{4}{3}\).

\[\frac{\cos x + \tan x}{\cot x} = \frac{\frac{4}{5} + \frac{3}{4}}{\frac{4}{3}} = \frac{31}{20}\cdot\frac{3}{4} = \frac{93}{80}.\]

Problem 3 Ratios from one ratio

a) In a right triangle, \(\tan x = \frac{2}{3}\). Find \(\dfrac{\sin x\cdot\cos x}{\cot x}\).

b) In a right triangle \(ABC\) the right angle is at \(A\), \(AC = 3\), \(BC = 5\) and \(m(\angle ACB) = x\). Find \(\dfrac{\sin x\cdot\cos x}{\tan x + \cot x}\).

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a) Opposite side 2, adjacent side 3, hypotenuse \(\sqrt{13}\). So \(\sin x = \frac{2}{\sqrt{13}}\), \(\cos x = \frac{3}{\sqrt{13}}\), \(\cot x = \frac{3}{2}\).

\[\frac{\sin x\cos x}{\cot x} = \frac{6}{13}\cdot\frac{2}{3} = \frac{4}{13}.\]

b) \(AB = \sqrt{5^2 - 3^2} = 4\). For the angle at \(C\): \(\sin x = \frac{AB}{BC} = \frac{4}{5}\), \(\cos x = \frac{AC}{BC} = \frac{3}{5}\).

Since \(\tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{1}{\sin x\cos x}\), the expression equals \((\sin x\cos x)^2 = \left(\frac{12}{25}\right)^2 = \frac{144}{625}\).

Problem 4 Isosceles triangle

In triangle \(ABC\), \(|AB| = |AC|\) and \(\sin C = \frac{3}{4}\). Find \(\cos B\), \(\tan B\) and \(\cot B\).

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The base angles of an isosceles triangle are equal and acute, so \(B = C\) and \(\sin B = \frac{3}{4}\).

\(\cos B = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}\).

\(\tan B = \frac{3/4}{\sqrt7/4} = \frac{3}{\sqrt{7}} = \frac{3\sqrt7}{7}\), \(\cot B = \frac{\sqrt7}{3}\).

Problem 5 Complementary angles

Find the unknown acute angle.

a) \(\cos x = \sin 7^\circ\) b) \(\cot y = \tan 25^\circ\) c) \(\sin m = \cos\frac{\pi}{9}\) d) \(\csc n = \sec\frac{\pi}{12}\)

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For acute angles: \(\sin\alpha = \cos(90^\circ - \alpha)\), \(\tan\alpha = \cot(90^\circ - \alpha)\), \(\sec\alpha = \csc(90^\circ - \alpha)\).

a) \(x = 90^\circ - 7^\circ = 83^\circ\)
b) \(y = 90^\circ - 25^\circ = 65^\circ\)
c) \(m = \frac{\pi}{2} - \frac{\pi}{9} = \frac{7\pi}{18}\)
d) \(n = \frac{\pi}{2} - \frac{\pi}{12} = \frac{5\pi}{12}\)

Problem 6 Complementary angles

Simplify.

a) \(\tan\frac{\pi}{9}\cdot\tan\frac{7\pi}{18}\) b) \(\cos^2 15^\circ + \cos^2 75^\circ\) c) \(-\sin^2 20^\circ - \sin^2 70^\circ\)

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a) \(\frac{7\pi}{18} = \frac{\pi}{2} - \frac{\pi}{9}\), so \(\tan\frac{7\pi}{18} = \cot\frac{\pi}{9}\) and the product is \(\tan\frac{\pi}{9}\cot\frac{\pi}{9} = 1\).
b) \(\cos 75^\circ = \sin 15^\circ\), so the sum is \(\cos^2 15^\circ + \sin^2 15^\circ = 1\).
c) \(\sin 70^\circ = \cos 20^\circ\), so we get \(-(\sin^2 20^\circ + \cos^2 20^\circ) = -1\).

Problem 7 Special angles

Calculate.

a) \(\sin 30^\circ + \cos 60^\circ\) b) \(\tan 60^\circ - \cot 30^\circ\) c) \(\sec 45^\circ - \csc 45^\circ\)

d) \(\sin 60^\circ\cdot\cos 30^\circ\) e) \(\tan 45^\circ + \sin 45^\circ\) f) \(\cos 60^\circ + \sec 60^\circ\)

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Values: \(\sin 30^\circ = \cos 60^\circ = \frac12\), \(\sin 60^\circ = \cos 30^\circ = \frac{\sqrt3}{2}\), \(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt2}{2}\), \(\tan 60^\circ = \cot 30^\circ = \sqrt3\), \(\tan 45^\circ = 1\).

a) \(\frac12 + \frac12 = 1\)
b) \(\sqrt3 - \sqrt3 = 0\)
c) \(\sqrt2 - \sqrt2 = 0\)
d) \(\frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2} = \frac34\)
e) \(1 + \frac{\sqrt2}{2}\)
f) \(\frac12 + 2 = \frac52\)

Problem 8 Lengths as ratios

In triangle \(ABC\) the right angle is at \(A\), \(AD\) is the altitude to \(BC\), \(AB = 2\) and \(m(\angle ABC) = \theta\). Write \(|BD|\), \(|AD|\) and \(|AC|\) using trigonometric ratios of \(\theta\).

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In the right triangle \(ABD\) (right angle at \(D\)) the hypotenuse is \(AB = 2\):

\(|BD| = AB\cos\theta = 2\cos\theta\), \(|AD| = AB\sin\theta = 2\sin\theta\).

In the right triangle \(ABC\) (right angle at \(A\)): \(\tan\theta = \frac{AC}{AB}\), so \(|AC| = 2\tan\theta\).

Problem 9 Simplifying expressions

Simplify.

a) \(\cos x\cdot\tan x\cdot\csc x\) b) \((\tan x + \cot x)\sin x\cos x\) c) \(\dfrac{\sin x\cos x}{\cot x} + \dfrac{\cos x}{\sec x}\)

d) \(\dfrac{1 + \cos x}{1 + \sec x}\) e) \(\dfrac{1 + \sin x}{\cos x} - \dfrac{\cos x}{1 - \sin x}\) f) \(\dfrac{(\sin x + \cos x)^2 - 1}{(\sin x - \cos x)^2 - 1}\) g) \((1 - \cos^2 x)(1 + \cot^2 x)\)

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a) \(\cos x\cdot\frac{\sin x}{\cos x}\cdot\frac{1}{\sin x} = 1\)
b) \(\left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right)\sin x\cos x = \sin^2 x + \cos^2 x = 1\)
c) \(\frac{\sin x\cos x}{\cos x/\sin x} + \cos x\cdot\cos x = \sin^2 x + \cos^2 x = 1\)
d) \(1 + \sec x = \frac{\cos x + 1}{\cos x}\), so the fraction is \((1 + \cos x)\cdot\frac{\cos x}{1 + \cos x} = \cos x\)
e) Common denominator \(\cos x(1 - \sin x)\): the numerator is \((1 + \sin x)(1 - \sin x) - \cos^2 x = 1 - \sin^2 x - \cos^2 x = 0\). The result is \(0\).
f) \((\sin x \pm \cos x)^2 - 1 = \pm 2\sin x\cos x\), so the fraction is \(\frac{2\sin x\cos x}{-2\sin x\cos x} = -1\)
g) \(\sin^2 x\cdot\csc^2 x = 1\)

Problem 10 Verifying identities

Verify the identities.

a) \(\cot^2 x - \cos^2 x = \cot^2 x\cdot\cos^2 x\)

b) \(\dfrac{\sin x}{1 + \cos x} + \dfrac{1 + \cos x}{\sin x} = 2\csc x\)

c) \(\dfrac{\cos x}{1 - \sin x} - \tan x = \sec x\)

d) \(\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} = 2\sec^2 x\)

e) \(\sec^2 x + \csc^2 x = \sec^2 x\cdot\csc^2 x\)

Show solution

a) \(\cot^2 x - \cos^2 x = \cos^2 x\left(\frac{1}{\sin^2 x} - 1\right) = \cos^2 x\cdot\frac{\cos^2 x}{\sin^2 x} = \cos^2 x\cot^2 x\).

b) Common denominator: \(\frac{\sin^2 x + (1 + \cos x)^2}{\sin x(1 + \cos x)} = \frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{\sin x(1 + \cos x)} = \frac{2(1 + \cos x)}{\sin x(1 + \cos x)} = \frac{2}{\sin x} = 2\csc x\).

c) \(\frac{\cos x}{1 - \sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin x + \sin^2 x}{\cos x(1 - \sin x)} = \frac{1 - \sin x}{\cos x(1 - \sin x)} = \frac{1}{\cos x} = \sec x\).

d) \(\frac{(1 + \sin x) + (1 - \sin x)}{1 - \sin^2 x} = \frac{2}{\cos^2 x} = 2\sec^2 x\).

e) \(\frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x\cos^2 x} = \frac{1}{\cos^2 x}\cdot\frac{1}{\sin^2 x} = \sec^2 x\csc^2 x\).