Essential trigonometric identities: Pythagorean, ratio, complementary and double angle

Use the double angle identities $\cos 2\theta = 1 - 2 \sin^2 \theta$ and $\sin 2\theta = 2 \sin \theta \cos \theta$.

$\frac{1 - \cos 2\theta}{\sin 2\theta} = \frac{1 - (1 - 2 \sin^2 \theta)}{2 \sin \theta \cos \theta} = \frac{2 \sin^2 \theta}{2 \sin \theta \cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$.

Markers reward the choice of the correct double angle forms, the cancellation, and a final line that is clearly the right-hand side.

In the second quadrant $\sin > 0$ and $\cos < 0$.

Pythagorean identity: $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$, so $\cos \theta = -\frac{4}{5}$ (negative in Q2).

$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{3}{5} \cdot \left( -\frac{4}{5} \right) = -\frac{24}{25}$.

Markers expect the correct sign of $\cos \theta$ from the quadrant, the double angle formula, and the exact answer.

Recognise the Pythagorean identity in the denominator. From $\sin^2\theta + \cos^2\theta = 1$, rearranging gives $1 - \cos^2\theta = \sin^2\theta$:

Marker's note: one mark for replacing $1 - \cos^2\theta$ with $\sin^2\theta$ via the Pythagorean identity, one for the simplified value $1$. Leaving the answer as $\dfrac{\sin^2\theta}{\sin^2\theta}$ without cancelling does not earn the second mark.

Find $\sin\theta$ with the Pythagorean identity. In the first quadrant $\sin\theta > 0$:

Apply the double angle formula.

Marker's note: one mark for $\sin\theta = \frac{12}{13}$ (positive from Q1), one for $\sin 2\theta = \frac{120}{169}$ via $2\sin\theta\cos\theta$. Forgetting the factor of $2$ is the usual slip.

Work from the left-hand side only. Replace $\cos^2\theta$ using the Pythagorean identity $\cos^2\theta = 1 - \sin^2\theta$, which factors as a difference of two squares:

Cancel the common factor.

As required.

Marker's note: one mark for $\cos^2\theta = 1 - \sin^2\theta$, one for factoring $1 - \sin^2\theta = (1 - \sin\theta)(1 + \sin\theta)$, one for cancelling to reach the RHS. Juggling both sides at once scores zero for method.

Convert the left-hand side to $\sin$ and $\cos$.

Put over a common denominator and use the Pythagorean identity.

Introduce the double angle form. Since $\sin 2\theta = 2\sin\theta\cos\theta$, we have $\sin\theta\cos\theta = \dfrac{\sin 2\theta}{2}$:

As required.

Marker's note: one mark for converting to $\sin$ and $\cos$ over a common denominator, one for simplifying the numerator to $1$ via $\sin^2\theta + \cos^2\theta = 1$, one for the double angle substitution to reach $\dfrac{2}{\sin 2\theta}$. Starting from the RHS is equally valid but must still work down one side only.

Set $y = 0$ and substitute the double angle form.

Solve for $\sin x$ over the domain. Since $0 \le x \le \pi$, $\sin x \ge 0$, so we take the positive root:

Confirm against the graph. These are exactly the two $x$-intercepts described, at $x = \dfrac{\pi}{4}$ and $x = \dfrac{3\pi}{4}$, so the algebra matches the picture.

Marker's note: one mark for reducing to $\sin^2 x = \frac{1}{2}$, one for both solutions $x = \frac{\pi}{4}, \frac{3\pi}{4}$ in the domain (rejecting the negative root because $\sin x \ge 0$ on $[0,\pi]$), one for stating they match the graph's intercepts. Working in degrees, or dropping the second solution, loses marks.

Find $\sin\theta$ and $\cos\theta$ with the correct quadrant signs. In the third quadrant both $\sin\theta < 0$ and $\cos\theta < 0$. From $\tan\theta = \dfrac{3}{4}$, a reference triangle has opposite $3$, adjacent $4$, hypotenuse $5$, so

Apply the double angle formulas.

Marker's note: one mark for the reference triangle values $\frac{3}{5}$ and $\frac{4}{5}$, one for fixing both signs negative in Q3, one for $\sin 2\theta = \frac{24}{25}$, one for $\cos 2\theta = \frac{7}{25}$. Note $\sin 2\theta > 0$ even though $\theta$ is in Q3, because the two negatives multiply to a positive.

Prove the identity, working from the left-hand side. Substitute the double angle forms $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$ (chosen so the $+1$ cancels):

Cancel the common $2\cos\theta$:

As required.

Hence solve the equation. The identity turns the equation into

Over $0 \le \theta \le \pi$, tangent takes the value $\dfrac{1}{\sqrt{3}}$ at

The other solution of $\tan\theta = \frac{1}{\sqrt{3}}$ in the interval would be $\theta = \frac{\pi}{6} + \pi$, which exceeds $\pi$, and $\theta = \frac{\pi}{2}$ is excluded because $\cos\theta = 0$ there, so the only solution is $\theta = \dfrac{\pi}{6}$.

Marker's note: two marks for the proof (one for the double angle substitutions with the form of $\cos 2\theta$ that cancels, one for cancelling to $\tan\theta$), one for using it to reach $\tan\theta = \frac{1}{\sqrt{3}}$, one for $\theta = \frac{\pi}{6}$, and one for justifying that no other value in $[0,\pi]$ qualifies (checking the restriction $\cos\theta \neq 0$). Omitting the domain check caps the final mark.

Show the identity by factoring a difference of two squares.

The second factor is $1$ by the Pythagorean identity, and the first is a form of $\cos 2\theta$:

As required.

Hence integrate the simpler expression. Replacing the integrand with $\cos 2\theta$:

Evaluate at the limits.

Marker's note: one mark for factoring $\cos^4\theta - \sin^4\theta$ as a difference of squares, one for using $\cos^2\theta + \sin^2\theta = 1$ and identifying $\cos 2\theta$, one for the antiderivative $\frac{1}{2}\sin 2\theta$, one for correct substitution of the limits in radians, one for the exact value $\frac{1}{2}$. Integrating $\cos^4\theta - \sin^4\theta$ directly without simplifying first is the trap; the "show that" hands you the shortcut.