IB Mathematics Analysis and Approaches Topic SL…

Why Identities Help: Rewriting Without Solving

A trigonometric identity is an equation that is true for every angle, so it lets you replace one expression with an equal one whenever that makes a problem simpler.
The real power is that identities let you find a value, such as $\tan\theta$ or $\sin 2\theta$, without ever finding the angle itself, which is exactly what exam questions ask for.
The two identities in this section, the Pythagorean identity and the double angle identities, are the workhorses for the equations you meet in SL 3.8.

The Pythagorean Identity

For every angle, $\cos^2\theta + \sin^2\theta = 1$, given in the formula booklet.
Where it comes from: on the unit circle the point $(\cos\theta, \sin\theta)$ lies on the circle $x^2 + y^2 = 1$, so substituting the coordinates gives the identity directly.
The useful rearrangements are $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$, which let you swap between sine and cosine.
Note that $\sin^2\theta$ means $(\sin\theta)^2$, the whole ratio squared, not the sine of $\theta^2$.

Example question

Given that $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find $\cos\theta$.

Solution

Rearrange the identity: $\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$.
Take the square root: $\cos\theta = \pm\frac{4}{5}$.
Since $\theta$ is acute, cosine is positive, so $\cos\theta = \frac{4}{5}$.

Finding One Ratio From Another

Combining the Pythagorean identity with the definition $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ lets you find any ratio from any other.
The quadrant, or a statement such as acute or obtuse, tells you which sign to keep after the square root.

Example question

Given that $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find $\tan\theta$ without finding $\theta$.

Solution

From the previous result, $\cos\theta = \frac{4}{5}$.
Use $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{3}{5}}{\frac{4}{5}}$.
Simplify: $$\tan\theta = \frac{3}{4}$$

Double Angle Identities

For sine, $\sin 2\theta = 2\sin\theta\cos\theta$, given in the formula booklet.
For cosine, $\cos 2\theta = \cos^2\theta - \sin^2\theta$, which the booklet also writes as $2\cos^2\theta - 1$ and $1 - 2\sin^2\theta$.
The three forms of $\cos 2\theta$ are all equal; you get from one to the next by substituting the Pythagorean identity, and you pick whichever form matches what you already know.
A common trap: $\cos 2\theta$ is not the same as $2\cos\theta$, and $\sin 2\theta$ is not $2\sin\theta$.

Example question

Given that $\cos x = \dfrac{3}{4}$ and $x$ is acute, find $\sin 2x$ without finding $x$.

Solution

First find $\sin x$: $\sin^2 x = 1 - \cos^2 x = 1 - \frac{9}{16} = \frac{7}{16}$, so $\sin x = \frac{\sqrt{7}}{4}$ (positive, as $x$ is acute).
Apply the double angle identity: $$\sin 2x = 2\sin x\cos x = 2 \cdot \frac{\sqrt{7}}{4} \cdot \frac{3}{4}$$
Simplify: $$\sin 2x = \frac{3\sqrt{7}}{8}$$

Common Mistake

Reading $\sin^2\theta$ as $\sin(\theta^2)$; it means $(\sin\theta)^2$.
Writing $\cos 2\theta = 2\cos\theta$ or $\sin 2\theta = 2\sin\theta$; the double angle identities are not simple doubling.
Dropping the plus-or-minus after taking a square root, then not checking the quadrant to fix the sign.
Choosing the wrong form of $\cos 2\theta$ for the information given, which makes the algebra longer than it needs to be.

Active recall

State the Pythagorean identity and derive it from the unit circle.
Given $\sin\theta$ and the quadrant of $\theta$, how do you find $\cos\theta$?
Write the double angle identity for $\sin 2\theta$.
Give the three equivalent forms of $\cos 2\theta$.
Why is $\cos 2\theta$ not equal to $2\cos\theta$?

Why Identities Help: Rewriting Without Solving

A trigonometric identity is an equation that is true for every angle, so it lets you replace one expression with an equal one whenever that makes a problem simpler.
The real power is that identities let you find a value, such as $\tan\theta$ or $\sin 2\theta$, without ever finding the angle itself, which is exactly what exam questions ask for.
The two identities in this section, the Pythagorean identity and the double angle identities, are the workhorses for the equations you meet in SL 3.8.

The Pythagorean Identity

For every angle, $\cos^2\theta + \sin^2\theta = 1$, given in the formula booklet.
Where it comes from: on the unit circle the point $(\cos\theta, \sin\theta)$ lies on the circle $x^2 + y^2 = 1$, so substituting the coordinates gives the identity directly.
The useful rearrangements are $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$, which let you swap between sine and cosine.
Note that $\sin^2\theta$ means $(\sin\theta)^2$, the whole ratio squared, not the sine of $\theta^2$.

Example question

Given that $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find $\cos\theta$.

Solution

Rearrange the identity: $\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$.
Take the square root: $\cos\theta = \pm\frac{4}{5}$.
Since $\theta$ is acute, cosine is positive, so $\cos\theta = \frac{4}{5}$.

Finding One Ratio From Another

Combining the Pythagorean identity with the definition $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ lets you find any ratio from any other.
The quadrant, or a statement such as acute or obtuse, tells you which sign to keep after the square root.

Example question

Given that $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find $\tan\theta$ without finding $\theta$.

Solution

From the previous result, $\cos\theta = \frac{4}{5}$.
Use $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{3}{5}}{\frac{4}{5}}$.
Simplify: $$\tan\theta = \frac{3}{4}$$

Double Angle Identities

For sine, $\sin 2\theta = 2\sin\theta\cos\theta$, given in the formula booklet.
For cosine, $\cos 2\theta = \cos^2\theta - \sin^2\theta$, which the booklet also writes as $2\cos^2\theta - 1$ and $1 - 2\sin^2\theta$.
The three forms of $\cos 2\theta$ are all equal; you get from one to the next by substituting the Pythagorean identity, and you pick whichever form matches what you already know.
A common trap: $\cos 2\theta$ is not the same as $2\cos\theta$, and $\sin 2\theta$ is not $2\sin\theta$.

Example question

Given that $\cos x = \dfrac{3}{4}$ and $x$ is acute, find $\sin 2x$ without finding $x$.

Solution

First find $\sin x$: $\sin^2 x = 1 - \cos^2 x = 1 - \frac{9}{16} = \frac{7}{16}$, so $\sin x = \frac{\sqrt{7}}{4}$ (positive, as $x$ is acute).
Apply the double angle identity: $$\sin 2x = 2\sin x\cos x = 2 \cdot \frac{\sqrt{7}}{4} \cdot \frac{3}{4}$$
Simplify: $$\sin 2x = \frac{3\sqrt{7}}{8}$$

Common Mistake

Reading $\sin^2\theta$ as $\sin(\theta^2)$; it means $(\sin\theta)^2$.
Writing $\cos 2\theta = 2\cos\theta$ or $\sin 2\theta = 2\sin\theta$; the double angle identities are not simple doubling.
Dropping the plus-or-minus after taking a square root, then not checking the quadrant to fix the sign.
Choosing the wrong form of $\cos 2\theta$ for the information given, which makes the algebra longer than it needs to be.

Active recall

State the Pythagorean identity and derive it from the unit circle.
Given $\sin\theta$ and the quadrant of $\theta$, how do you find $\cos\theta$?
Write the double angle identity for $\sin 2\theta$.
Give the three equivalent forms of $\cos 2\theta$.
Why is $\cos 2\theta$ not equal to $2\cos\theta$?

IB Mathematics Analysis and Approaches SL 3.6—pythagorean Identity, Double Angles Notes

IB Mathematics Analysis and Approaches SL 3.6—pythagorean Identity, Double Angles Notes

Why Identities Help: Rewriting Without Solving

The Pythagorean Identity

Finding One Ratio From Another

Double Angle Identities

Why Identities Help: Rewriting Without Solving

The Pythagorean Identity

Finding One Ratio From Another

Double Angle Identities

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches SL 3.6—pythagorean Identity, Double Angles Notes

IB Mathematics Analysis and Approaches SL 3.6—pythagorean Identity, Double Angles Notes

Why Identities Help: Rewriting Without Solving

The Pythagorean Identity

Finding One Ratio From Another

Double Angle Identities

Why Identities Help: Rewriting Without Solving

The Pythagorean Identity

Finding One Ratio From Another

Double Angle Identities