KindiakMath

Trigonometric Identities

In these exercises, you may use the following addition formulae for \sin and \cos: for any A, B \in \mathbb R,

\begin{aligned} \sin(A+B) &= \sin(A) \cos(B) + \cos(A) \sin(B), \\ \cos(A+B) &= \cos(A) \cos(B) - \sin(A) \sin(B). \end{aligned}

Problem 1. Prove the following negative identities: for any A \in \mathbb R, whenever the right-hand side is well-defined,

\sin(-A) = -\sin(A),\quad \cos(-A) = \cos(A),\quad \tan(-A) = -\tan(A).

Deduce the subtraction formulae for \sin and \cos: for any A, B \in \mathbb R,

\begin{aligned} \sin(A-B) &= \sin(A) \cos(B) - \cos(A) \sin(B), \\ \cos(A-B) &= \cos(A) \cos(B) + \sin(A) \sin(B). \end{aligned}

(Click for Solution)

Solution. For the first two identities, we use the addition formulae to derive

\begin{aligned} 0 = \sin(-A + A) &= \sin(-A) \cos(A) + \cos(-A) \sin(A), \\ 1 = \cos(-A + A) &= \cos(-A) \cos(A) - \sin(-A) \sin(A). \end{aligned}

Solving simultaneous linear equations yields \sin(-A) = -\sin(A) and \cos(-A) = \cos(A). For the final identity,

\displaystyle \tan(-A) = \frac{\sin(-A)}{\cos(-A)} = \frac{-\sin(A)}{\cos(A)} = -\tan(A).

Replacing B with -B in the addition formulae,

\begin{aligned} \sin(A-B) &= \sin(A+(-B)) \\ &= \sin(A)\cos(-B) + \cos(A) \sin(-B)\\ &= \sin(A) \cos(B) - \cos(A) \sin(B). \end{aligned}

A similar calculation yields the cosine subtraction formula.

Problem 2. Prove the following double angle identities: for any A \in \mathbb R,

\sin(2A) = 2 \sin(A) \cos(A),\quad \cos(2A) =\cos^2(A) - \sin^2(A).

(Click for Solution)

Solution. Setting A = B,

\begin{aligned} \sin(2A) &= \sin(A+A) \\ &= \sin(A) \cos(A) + \cos(A) \sin(A) \\ &= 2 \sin(A) \cos(A), \\ \cos(2A) &= \cos(A+A) \\ &= \cos(A) \cos(A) - \sin(A) \sin(A) \\ &= \cos^2(A) - \sin^2(A). \end{aligned}

Problem 3. Prove that for any A \in \mathbb R,

\sin^2(A) + \cos^2(A) = 1.

Deduce that for any A \in \mathbb R,

\begin{aligned} \cos(2A) &=\cos^2(A) - \sin^2(A) \\ &= 2 \cos^2(A) - 1\\ &= 1 - 2\sin^2(A). \end{aligned}

(Click for Solution)

Solution. For the first identity, it suffices to prove the result for A \in (0, 2\pi], since

\sin(A + n \cdot 2\pi) = \sin(A),\quad \cos(A + n \cdot 2\pi) = \cos(A).

We have already proven the case A \in (0, \pi/4] using the vanilla Pythagorean theorem. For the case A \in (0, \pi/2], denote A_2 = A/2 and use Problem 2 to obtain

\begin{aligned} \sin^2(A) + \cos^2(A) &= \sin^2(2A_2) + \cos^2(2A_2) \\ &= (2 \sin(A_2) \cos(A_2))^2 + (\cos^2(A_2) - \sin^2(A_2))^2 \\ &= \sin^4(A_2) + 2 \sin^2(A_2) \cos^2(A_2) + \cos^4(A_2) \\ &= (\sin^2(A_2) + \cos^2(A_2))^2 = 1^2 = 1. \end{aligned}

By repeating the argument, we can “double” the result to hold for (0, \pi] and (0, 2\pi], as required. For the second identity, use Problem 2 and the first identity to obtain

\begin{aligned} \cos(2A) &= \cos^2(A) - \sin^2(A) \\ &= \cos^2(A) - (1 - \cos^2(A)) \\ &= 2 \cos^2(A) - 1 \\ &= 2(1 - \sin^2(A)) - 1 \\ &= 1 - 2 \sin^2(A). \end{aligned}

Problem 4. Define \tan(A) := \sin(A)/{\cos(A)} whenever the right-hand side is well-defined. Define the following reciprocal trigonometric functions whenever their right-hand sides are well-defined:

\displaystyle \sec(A) := \frac{1}{\cos(A)},\quad \csc(A) := \frac{1}{\sin(A)},\quad \cot(A) := \frac{1}{\tan(A)}.

Prove the following Pythagorean identities, whenever they are well-defined:

\begin{aligned} \tan^2(A) + 1 &= \sec^2(A), \\ 1 + \cot^2(A) &= \csc^2(A). \end{aligned}

(Click for Solution)

Solution. Using Problem 3,

\begin{aligned} \tan^2(A) + 1 &= \frac{\sin^2(A)}{\cos^2(A)} + 1 \\ &= \frac{\sin^2(A) + \cos^2(A)}{\cos^2(A)} \\ &= \frac 1{\cos^2(A)} = \sec^2(A). \end{aligned}

Similarly,

\begin{aligned} 1 + \cot^2(A) &= 1 + \frac{\cos^2(A)}{\sin^2(A)} + 1 \\ &= \frac{\sin^2(A) + \cos^2(A)}{\sin^2(A)} \\ &= \frac 1{\sin^2(A)} = \csc^2(A). \end{aligned}

Problem 5. Prove the following half angle identities: for any A \in \mathbb R,

\displaystyle \sin^2(A/2) =\frac {1 - \cos(A)}2, \quad \cos^2(A/2) =\frac {1 + \cos(A)}2.

(Click for Solution)

Solution. Using Problem 3,

\cos(A) = 2 \cos^2(A/2) - 1 = 1 - 2\sin^2(A/2).

Algebruh yields the desired result.

Problem 6. Prove the following triple angle identities: for any A \in \mathbb R,

\begin{aligned} \sin(3A) &=3 \sin(A) - 4 \sin^3(A), \\ \cos(3A) &= 4 \cos^3(A) - 3 \cos(A).\end{aligned}

(Click for Solution)

Solution. Using the addition formulae and Problem 3,

\begin{aligned} \sin(3A) &= \sin(A + 2A) \\ &= \sin(A) \cos(2A) + \cos(A) \sin(2A) \\ &= \sin(A) (1 - 2\sin^2(A)) + \cos(A) \cdot 2 \sin(A)\cos(A) \\ &= \sin(A) (1 - 2\sin^2(A)) + 2 \sin(A) (1 - \sin^2(A))\\ &= \sin(A) - 2\sin^3(A) + 2 \sin(A) - 2 \sin^3(A) \\ &= 3 \sin(A) - 4 \sin^3(A). \end{aligned}

Similarly,

\begin{aligned} \cos(3A) &= \cos(A + 2A) \\ &= \cos(A) \cos(2A) - \sin(A) \sin(2A) \\ &= \cos(A) (2\cos^2(A)-1) - \sin(A) \cdot 2 \sin(A)\cos(A) \\ &= \cos(A) (2\cos^2(A)-1) - 2 \cos(A) (1 - \cos^2(A))\\ &= 2\cos^3(A) - \cos(A) - 2 \cos(A) + 2 \cos^3(A) \\ &= 4 \cos^3(A) - 3 \cos(A). \end{aligned}

Problem 7. Prove the following tangent identities: for any A,B \in \mathbb R, whenever the right-hand side is well-defined,

\begin{aligned} \tan(A + B) &= \frac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}, \\ \tan(3A) &= \frac{3 \tan(A) - \tan^3(A) }{ 1 - 3 \tan^2(A) }.\end{aligned}

(Click for Solution)

Solution. Using the addition formulae,

\begin{aligned} \tan(A + B) &= \frac{\sin(A+B)}{\cos(A+B)} \\ &= \frac{\sin(A) \cos(B) + \cos(A) \sin(B)}{\cos(A) \cos(B) - \sin(A)\sin(B)} \\ &= \frac{\frac{\sin(A)}{\cos(A)} \cdot 1 + 1 \cdot \frac{\sin(B)}{\cos(B)} }{ 1 - \frac{\sin(A)}{\cos(A)} \cdot \frac{\sin(B)}{\cos(B)} } \\ &= \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}. \end{aligned}

Using Problems 4 and 6,

\begin{aligned} \tan(3A) &= \frac{\sin(3A)}{\cos(3A)} \\ &= \frac{3 \sin(A) - 4 \sin^3(A)}{4 \cos^3(A) - 3\cos(A)} \\ &= \frac{3 \tan(A) \sec^2(A) - 4 \tan^3(A)}{4 - 3 \sec^2(A)} \\ &= \frac{3 \tan(A) (1 + \tan^2(A)) - 4 \tan^3(A)}{4 - 3 (1 + \tan^2(A))} \\ &= \frac{ (3 \tan(A) + 3 \tan^3(A)) - 4 \tan^3(A)}{4 - (3 + 3\tan^2(A))} \\ &= \frac{ 3 \tan(A) - \tan^3(A) }{1 - 3\tan^2(A)}. \end{aligned}

—Joel Kindiak, 4 Jun 25, 1918H

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joelkindiak

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