Trigonometry Problems & Questions with Solutions

Prove the identity: $$ \tan^2(x) - \sin^2(x) = \tan^2(x) \sin^2(x) $$

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Solution:

We start with the left-hand side of the given identity.

Use the identity $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$ to write:

$$ \tan^2(x) - \sin^2(x) = \left(\dfrac{\sin(x)}{\cos(x)}\right)^2 - \sin^2(x) $$

$$ = \dfrac{\sin^2(x)}{\cos^2(x)} - \sin^2(x) $$

Find a common denominator:

$$ = \dfrac{\sin^2(x) - \cos^2(x) \sin^2(x)}{\cos^2(x)} $$

Factor out $\sin^2(x)$ from the numerator:

$$ = \dfrac{\sin^2(x) \left(1 - \cos^2(x)\right)}{\cos^2(x)} $$

Apply the Pythagorean identity $1 - \cos^2(x) = \sin^2(x)$:

$$ = \dfrac{\sin^2(x) \sin^2(x)}{\cos^2(x)} $$

$$ = \sin^2(x) \tan^2(x) $$

This is exactly equal to the right-hand side of the given identity.

Prove the identity: $$ \dfrac{1 + \cos(x) + \cos(2x)}{\sin(x) + \sin(2x)} = \cot(x) $$

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Solution:

Use the double angle identities $\cos(2x) = 2 \cos^2(x) - 1$ and $\sin(2x) = 2 \sin(x) \cos(x)$ in the left-hand side of the given identity:

$$ \dfrac{1 + \cos(x) + \cos(2x)}{\sin(x) + \sin(2x)} $$

$$ = \dfrac{1 + \cos(x) + 2 \cos^2(x) - 1}{\sin(x) + 2 \sin(x) \cos(x)} $$

Simplify the numerator by canceling the $1$ and $-1$:

$$ = \dfrac{\cos(x) + 2 \cos^2(x)}{\sin(x) + 2 \sin(x) \cos(x)} $$

Factor $\cos(x)$ from the numerator and $\sin(x)$ from the denominator:

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

Cancel the common binomial factor $(1 + 2 \cos(x))$:

$$ = \dfrac{\cos(x)}{\sin(x)} $$

$$ = \cot(x) $$

This is equal to the right-hand side.

Prove the identity: $$ 4 \sin(x) \cos(x) = \dfrac{\sin(4x)}{\cos(2x)} $$

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Solution:

Use the double angle identity $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$. By substituting $\theta = 2x$, we can rewrite $\sin(4x)$ as $2 \sin(2x) \cos(2x)$.

Substitute this into the right-hand side of the given identity:

$$ \dfrac{\sin(4x)}{\cos(2x)} = \dfrac{2 \sin(2x) \cos(2x)}{\cos(2x)} $$

Cancel $\cos(2x)$:

$$ = 2 \sin(2x) $$

Apply the double angle identity again for $\sin(2x)$:

$$ = 2 \times [2 \sin(x) \cos(x)] $$

$$ = 4 \sin(x) \cos(x) $$

This is equal to the left-hand side.

Solve the trigonometric equation:

$$ \sin(x) + \sin\left(\dfrac{x}{2}\right) = 0 \quad \text{for} \quad 0 \le x \le 2\pi $$

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Solution:

Use the identity $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$ to rewrite $\sin(x)$ by setting $\theta = \dfrac{x}{2}$:

$$ \sin(x) = \sin\left(2 \left( \dfrac{x}{2} \right)\right) = 2 \sin\left(\dfrac{x}{2}\right) \cos\left(\dfrac{x}{2}\right) $$

Substitute this into the given equation:

$$ 2 \sin\left(\dfrac{x}{2}\right) \cos\left(\dfrac{x}{2}\right) + \sin\left(\dfrac{x}{2}\right) = 0 $$

Factor out $\sin\left(\dfrac{x}{2}\right)$:

$$ \sin\left(\dfrac{x}{2}\right) \left( 2 \cos\left(\dfrac{x}{2}\right) + 1 \right) = 0 $$

This gives two separate equations to solve:

a) $\sin\left(\dfrac{x}{2}\right) = 0$

The solutions for the angle are $\dfrac{x}{2} = 0$ and $\dfrac{x}{2} = \pi$.

Solve for $x$ to obtain: $x = 0$ and $x = 2\pi$.

b) $2 \cos\left(\dfrac{x}{2}\right) + 1 = 0$

This may be written as $\cos\left(\dfrac{x}{2}\right) = -\dfrac{1}{2}$.

The solutions for the angle are $\dfrac{x}{2} = \dfrac{2\pi}{3}$ and $\dfrac{x}{2} = \dfrac{4\pi}{3}$.

Solve for $x$ to obtain: $x = \dfrac{4\pi}{3}$ and $x = \dfrac{8\pi}{3}$.

Note that $\dfrac{8\pi}{3}$ is greater than $2\pi$ and falls outside the given interval, so it is rejected.

Final Solutions: $\left\{ 0, \dfrac{4\pi}{3}, 2\pi \right\}$

Solve the trigonometric equation:

$$ (2\sin(x) - 1)(\tan(x) - 1) = 0 \quad \text{for} \quad 0 \le x \le 2\pi $$

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Solution:

The given equation is already in factored form, which leads directly to two equations to solve:

$$ 2\sin(x) - 1 = 0 \quad \text{and} \quad \tan(x) - 1 = 0 $$

The above equations may be written as:

$$ \sin(x) = \dfrac{1}{2} \quad \text{and} \quad \tan(x) = 1 $$

The solutions of the equation $\sin(x) = \dfrac{1}{2}$ in the interval $[0, 2\pi]$ are:

$$ x = \dfrac{\pi}{6} \quad \text{and} \quad x = \dfrac{5\pi}{6} $$

The solutions of the equation $\tan(x) = 1$ in the interval $[0, 2\pi]$ are:

$$ x = \dfrac{\pi}{4} \quad \text{and} \quad x = \dfrac{5\pi}{4} $$

Final Solutions: $\left\{ \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{5\pi}{6}, \dfrac{5\pi}{4} \right\}$

Solve the trigonometric equation:

$$ \cos(2x) \cos(x) - \sin(2x) \sin(x) = 0 \quad \text{for} \quad 0 \le x \le 2\pi $$

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Solution:

Use the formula for $\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ to condense the expression. Here, let $A = 2x$ and $B = x$:

$$ \cos(2x + x) = \cos(2x) \cos(x) - \sin(2x) \sin(x) $$

Hence the given equation may be written as:

$$ \cos(3x) = 0 $$

Since $x \in [0, 2\pi]$, the domain for $3x$ is $[0, 6\pi]$. Solve the equation for $3x$ to obtain all zeros of the cosine function in this extended interval:

$$ 3x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2}, \dfrac{7\pi}{2}, \dfrac{9\pi}{2}, \dfrac{11\pi}{2} $$

Divide each by $3$ to solve for $x$:

Final Solutions: $\left\{ \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{3\pi}{2}, \dfrac{11\pi}{6} \right\}$

Simplify the trigonometric expression:

$$ \dfrac{\sin(2x) - \cos(x)}{\cos(2x) + \sin(x) - 1} $$

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Solution:

Use the identities $\sin(2x) = 2 \sin(x) \cos(x)$ and $\cos(2x) = 1 - 2 \sin^2(x)$ to rewrite the expression:

$$ \dfrac{2 \sin(x) \cos(x) - \cos(x)}{1 - 2 \sin^2(x) + \sin(x) - 1} $$

Simplify the denominator by canceling $1$ and $-1$:

$$ = \dfrac{2 \sin(x) \cos(x) - \cos(x)}{-2 \sin^2(x) + \sin(x)} $$

Factor $\cos(x)$ from the numerator and $\sin(x)$ from the denominator:

$$ = \dfrac{\cos(x)(2 \sin(x) - 1)}{\sin(x)(-2 \sin(x) + 1)} $$

Notice that $(2 \sin(x) - 1)$ and $(-2 \sin(x) + 1)$ are opposites (they multiply by $-1$ to equal each other). Cancel them out, leaving a $-1$:

$$ = - \dfrac{\cos(x)}{\sin(x)} $$

$$ = - \cot(x) $$

Prove that:

$$ \sin(105^\circ) = \dfrac{\sqrt{6} + \sqrt{2}}{4} $$

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Solution:

$105^\circ$ may be written as the sum of two special angles:

$$ 105^\circ = 60^\circ + 45^\circ $$

Hence:

$$ \sin(105^\circ) = \sin(60^\circ + 45^\circ) $$

Use the identity $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$ to write:

$$ \sin(105^\circ) = \sin(60^\circ)\cos(45^\circ) + \cos(60^\circ)\sin(45^\circ) $$

Use the table of special angles for exact values:

$$ = \left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right) + \left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right) $$

Multiply the fractions:

$$ = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} $$

$$ = \dfrac{\sqrt{6} + \sqrt{2}}{4} $$

If $\sin(x) = \dfrac{2}{5}$ and $x$ is an acute angle, find the exact values of:

1. $\cos(2x)$
2. $\cos(4x)$
3. $\sin(2x)$
4. $\sin(4x)$

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Solution:

If $\sin(x) = \dfrac{2}{5}$, then using the Pythagorean identity $\cos(x) = \sqrt{1 - \sin^2(x)}$:

$$ \cos(x) = \sqrt{1 - \left(\dfrac{2}{5}\right)^2} = \sqrt{1 - \dfrac{4}{25}} = \sqrt{\dfrac{21}{25}} = \dfrac{\sqrt{21}}{5} $$

a) Find $\cos(2x)$:

Use the identity $\cos(2x) = 1 - 2\sin^2(x)$:

$$ \cos(2x) = 1 - 2\left(\dfrac{4}{25}\right) = 1 - \dfrac{8}{25} = \dfrac{17}{25} $$

c) Find $\sin(2x)$ first (needed for parts b and d):

Use the identity $\sin(2x) = 2 \sin(x) \cos(x)$:

$$ \sin(2x) = 2 \left(\dfrac{2}{5}\right) \left(\dfrac{\sqrt{21}}{5}\right) = \dfrac{4\sqrt{21}}{25} $$

b) Find $\cos(4x)$:

Use the identity $\cos(4x) = 2\cos^2(2x) - 1$:

$$ \cos(4x) = 2\left(\dfrac{17}{25}\right)^2 - 1 = 2\left(\dfrac{289}{625}\right) - 1 = \dfrac{578}{625} - \dfrac{625}{625} = -\dfrac{47}{625} $$

d) Find $\sin(4x)$:

Use the identity $\sin(4x) = \sin(2(2x)) = 2 \sin(2x) \cos(2x)$:

$$ \sin(4x) = 2 \left(\dfrac{4\sqrt{21}}{25}\right) \left(\dfrac{17}{25}\right) = \dfrac{136\sqrt{21}}{625} $$

Find the length of side $AB$ in the figure below. Round your answer to 3 significant digits.

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Solution:

Note that triangle $DAC$ is isosceles. Therefore, if we draw the perpendicular from $D$ to $AC$, it will cut $AC$ into two equal halves and bisect angle $D$.

Using the right triangle whose hypotenuse is $AD$ ($10$ units) and angle is $\dfrac{70^\circ}{2} = 35^\circ$, we can write:

$$ \sin(35^\circ) = \dfrac{\dfrac{1}{2}AC}{10} \implies \dfrac{1}{2}AC = 10 \sin(35^\circ) $$

Which gives:

$$ AC = 20 \sin(35^\circ) $$

Next, note that the two internal angles $B$ and $C$ of triangle $ABC$ add up to $90^\circ$ and therefore the third angle of triangle $ABC$ (Angle $A$) is a right angle.

We can therefore write:

$$ \tan(32^\circ) = \dfrac{AB}{AC} $$

Which gives:

$$ AB = AC \tan(32^\circ) $$

Substitute the value of $AC$ into the equation:

$$ AB = 20 \sin(35^\circ)\tan(32^\circ) \approx 7.17 \quad \text{(rounded to 3 significant digits)} $$

Solve the equation for all real values of $x \in [0, 2\pi]$:

$$ 2\sin^2(x) + 3\cos(x) = 0 $$

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Solution:

Use the Pythagorean identity $\sin^2(x) = 1 - \cos^2(x)$ to express everything in terms of cosine:

$$ 2(1 - \cos^2(x)) + 3\cos(x) = 0 $$

$$ 2 - 2\cos^2(x) + 3\cos(x) = 0 $$

Multiply by $-1$ to make the leading term positive:

$$ 2\cos^2(x) - 3\cos(x) - 2 = 0 $$

Let $u = \cos(x)$ to reveal the quadratic form:

$$ 2u^2 - 3u - 2 = 0 \implies (2u + 1)(u - 2) = 0 $$

This yields two equations:

$$ \cos(x) = -\dfrac{1}{2} \quad \text{and} \quad \cos(x) = 2 $$

Because the range of the cosine function is $[-1, 1]$, the equation $\cos(x) = 2$ has no real solution.

We solve $\cos(x) = -\dfrac{1}{2}$ for $x$ in the domain $[0, 2\pi]$:

$$ x = \dfrac{2\pi}{3}, \ \dfrac{4\pi}{3} $$

Prove the following identity:

$$ \sin(3x) + \sin(x) = 2\sin(2x)\cos(x) $$

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Solution:

We can prove this by starting on the left-hand side and applying the Sum-to-Product identity:

$$ \sin(A) + \sin(B) = 2\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right) $$

Let $A = 3x$ and $B = x$. Substitute these into the formula:

$$ \sin(3x) + \sin(x) = 2\sin\left(\dfrac{3x+x}{2}\right)\cos\left(\dfrac{3x-x}{2}\right) $$

Simplify the fractions inside the trigonometric functions:

$$ = 2\sin\left(\dfrac{4x}{2}\right)\cos\left(\dfrac{2x}{2}\right) $$

$$ = 2\sin(2x)\cos(x) $$

The identity is proven.