Mathematics 9709 · Pure Mathematics 3

Expressing a sin θ + b cos θ as R sin(θ ± α) or R cos(θ ∓ α)

$a\sin\theta+b\cos\theta$ looks messy because sine and cosine are out of step.
But add two waves of the same period and you always get one wave of that period back: taller by R and slid along by α.
Here $R=\sqrt{a^2+b^2}$ and $\tan\alpha=\tfrac{b}{a}$ .

Fold it into one wave: $a\sin\theta+b\cos\theta=R\sin(\theta+\alpha)$ with $R=\sqrt{a^2+b^2}$ and $\tan\alpha=\tfrac{b}{a}$ ( $\alpha$ acute). The pay-off: the height is $R$ , so the max is $R$ and the min is $-R$ .

See it: two waves become one

3 sin x + 4 cos x = 5 sin(x + 53.13°)·R = √(9 + 16) = 5

The dashed black curve is the real sum $a\sin\theta+b\cos\theta$ ; the gold curve is $R\sin(\theta+\alpha)$ . They lie right on top of each other for every $a$ and $b$ .
The gold wave never leaves the band between $R$ and $-R$ : that is why those are the max and min.

Match coefficients: find R and α

Write the target, expand it, and match up the $\sin\theta$ and $\cos\theta$ coefficients:

$a\sin\theta+b\cos\theta=R\sin(\theta+\alpha)=R\cos\alpha\,\sin\theta+R\sin\alpha\,\cos\theta$

Matching the coefficients gives $R\cos\alpha=a$ and $R\sin\alpha=b$ : square and add gives R, divide gives α.

$R=\sqrt{a^2+b^2}\ (R>0),\qquad \tan\alpha=\tfrac{b}{a}\ (0^\circ<\alpha<90^\circ)$

Square and add works because $(R\cos\alpha)^2+(R\sin\alpha)^2=R^2$ : the $\alpha$ cancels by Pythagoras.
Dividing the two equations cancels the $R$ and leaves just $\tan\alpha$ .

When the question asks for R cos(θ ∓ α)

If the expression starts with cos, fold it into a cosine wave instead: same $R$ and $\alpha$ , only the target line changes.
The exam picks the form for you, so just match whichever it asks for.

$a\cos\theta\pm b\sin\theta=R\cos(\theta\mp\alpha),\qquad R=\sqrt{a^2+b^2},\ \tan\alpha=\tfrac{b}{a}$

Note the flipped sign inside the bracket: $a\cos\theta-b\sin\theta$ becomes $R\cos(\theta+\alpha)$ : the minus turns into a plus.
To match, expand $R\cos(\theta+\alpha)=R\cos\alpha\cos\theta-R\sin\alpha\sin\theta$ : the $\cos\theta$ coefficient is $R\cos\alpha$ and the $\sin\theta$ coefficient is $R\sin\alpha$ .

Worked examples

Worked example

Express $3\sin\theta+4\cos\theta$ as $R\sin(\theta+\alpha)$

Expand the target and match: $R\cos\alpha=3$ (the $\sin$ coefficient), $R\sin\alpha=4$ (the $\cos$ coefficient).
Square-and-add: $R=\sqrt{3^2+4^2}=\sqrt{25}=5$ .
Divide: $\tan\alpha=\tfrac43\Rightarrow\alpha=53.13^\circ$ (acute).

Answer

3\sin\theta+4\cos\theta=5\sin(\theta+53.13^\circ)

Worked example

Real question: express $5\sin x-3\cos x$ as $R\sin(x-\alpha)$ , then state max and min

9709/31 O/N 2021 Q2. It leads with sin and the question wants a minus inside, so the target is $R\sin(x-\alpha)$ .

Expand the target: $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$ , so $R\cos\alpha=5$ and $R\sin\alpha=3$ .
$R=\sqrt{5^2+3^2}=\sqrt{34}$ (exact), and $\tan\alpha=\tfrac35\Rightarrow\alpha=30.96^\circ$ .
A sine runs from $-1$ to $+1$ , so $5\sin x-3\cos x$ runs from $-\sqrt{34}$ to $+\sqrt{34}$ .

Answer

\sqrt{34}\sin(x-30.96^\circ);\ \max=\sqrt{34},\ \min=-\sqrt{34}

Worked example

Solve $2\sin\theta-3\cos\theta=\sqrt2$ for $0^\circ<\theta<360^\circ$

Fold the left side (lead with sin, minus inside → $R\sin(\theta-\alpha)$ ): $R=\sqrt{13},\ \tan\alpha=\tfrac32\Rightarrow\alpha=56.31^\circ$ , so the equation becomes $\sqrt{13}\sin(\theta-56.31^\circ)=\sqrt2$ .
Isolate the sine: $\sin(\theta-56.31^\circ)=\sqrt{\tfrac{2}{13}}=0.3922$ .
Solve for the bracket: $\theta-56.31^\circ=23.09^\circ$ or $180^\circ-23.09^\circ=156.91^\circ$ .
Add $56.31^\circ$ back to each: $\theta=79.4^\circ$ or $213.2^\circ$ .

Answer

\theta=79.4^\circ\quad\text{or}\quad \theta=213.2^\circ

Worked example

Express $2\cos\theta-\sin\theta$ as $R\cos(\theta+\alpha)$ , then give its max and min

It leads with cos, so target $R\cos(\theta+\alpha)=R\cos\alpha\cos\theta-R\sin\alpha\sin\theta$ . Match: $R\cos\alpha=2$ , $R\sin\alpha=1$ .
$R=\sqrt{4+1}=\sqrt5$ , $\tan\alpha=\tfrac12\Rightarrow\alpha=26.57^\circ$ .
A cosine runs from $-1$ to $+1$ , so the max is $\sqrt5$ (at $\theta=333.4^\circ$ ) and the min is $-\sqrt5$ (at $\theta=153.4^\circ$ ).

Answer

\sqrt5\cos(\theta+26.57^\circ);\ \max=\sqrt5,\ \min=-\sqrt5

Why the max is R and where it sits

Reading the answer straight off the folded form

Once you have $R\sin(\theta+\alpha)$ , every “greatest value” question is instant. A sine tops out at $+1$ , so the expression tops out at $R\times1=R$ . That happens when the bracket $\theta+\alpha=90^\circ$ , that is $\theta=90^\circ-\alpha$ . The same idea gives the smallest value $-R$ when the bracket is $270^\circ$ .

Where the marks go

Common mistake

Keep

\alpha

in the first quadrant (

0^\circ<\alpha<90^\circ

): take the acute

\tan^{-1}\tfrac{b}{a}

, not a second-quadrant value.

Common mistake

The

\sin\theta

coefficient is

R\cos\alpha

(not

R\sin\alpha

). Write the expansion out in full before matching, or the whole answer flips.

Common mistake

When the question says exact, leave

R

as a surd like

\sqrt{34}

: it is

\alpha

that you give to 2 dp.

Now you try

Show that $5\sec x+\tan x=4$ can be written as $R\cos(x+\alpha)=5$ , giving the exact $R$ and $\alpha$ to 2 dp $(0^\circ<\alpha<90^\circ)$ . (9709/33 Oct/Nov 2022 Q7)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2022 Q7)

Multiply through by $\cos x$ ( $\sec x=\tfrac1{\cos x}$ , $\tan x=\tfrac{\sin x}{\cos x}$ ): $5+\sin x=4\cos x$ , so $4\cos x-\sin x=5$ .

It starts with cos, so fold the left side: $R=\sqrt{4^2+1^2}=\sqrt{17},\ \tan\alpha=\tfrac14\Rightarrow\alpha=14.04^\circ$ :

Answer

\sqrt{17}\cos(x+14.04^\circ)=5

PreviousFurther trigonometric identities NextProduct and Quotient Rule

Expressing a sin θ + b cos θ as R sin(θ ± α) or R cos(θ ∓ α)

$a\sin\theta+b\cos\theta$ looks messy because sine and cosine are out of step.
But add two waves of the same period and you always get one wave of that period back: taller by R and slid along by α.
Here $R=\sqrt{a^2+b^2}$ and $\tan\alpha=\tfrac{b}{a}$ .

See it: two waves become one

3 sin x + 4 cos x = 5 sin(x + 53.13°)·R = √(9 + 16) = 5

The dashed black curve is the real sum $a\sin\theta+b\cos\theta$ ; the gold curve is $R\sin(\theta+\alpha)$ . They lie right on top of each other for every $a$ and $b$ .
The gold wave never leaves the band between $R$ and $-R$ : that is why those are the max and min.

Match coefficients: find R and α

Write the target, expand it, and match up the $\sin\theta$ and $\cos\theta$ coefficients:

$a\sin\theta+b\cos\theta=R\sin(\theta+\alpha)=R\cos\alpha\,\sin\theta+R\sin\alpha\,\cos\theta$

Matching the coefficients gives $R\cos\alpha=a$ and $R\sin\alpha=b$ : square and add gives R, divide gives α.

$R=\sqrt{a^2+b^2}\ (R>0),\qquad \tan\alpha=\tfrac{b}{a}\ (0^\circ<\alpha<90^\circ)$

Square and add works because $(R\cos\alpha)^2+(R\sin\alpha)^2=R^2$ : the $\alpha$ cancels by Pythagoras.
Dividing the two equations cancels the $R$ and leaves just $\tan\alpha$ .

When the question asks for R cos(θ ∓ α)

If the expression starts with cos, fold it into a cosine wave instead: same $R$ and $\alpha$ , only the target line changes.
The exam picks the form for you, so just match whichever it asks for.

$a\cos\theta\pm b\sin\theta=R\cos(\theta\mp\alpha),\qquad R=\sqrt{a^2+b^2},\ \tan\alpha=\tfrac{b}{a}$

Note the flipped sign inside the bracket: $a\cos\theta-b\sin\theta$ becomes $R\cos(\theta+\alpha)$ : the minus turns into a plus.
To match, expand $R\cos(\theta+\alpha)=R\cos\alpha\cos\theta-R\sin\alpha\sin\theta$ : the $\cos\theta$ coefficient is $R\cos\alpha$ and the $\sin\theta$ coefficient is $R\sin\alpha$ .

Worked examples

Worked example

Express $3\sin\theta+4\cos\theta$ as $R\sin(\theta+\alpha)$

Expand the target and match: $R\cos\alpha=3$ (the $\sin$ coefficient), $R\sin\alpha=4$ (the $\cos$ coefficient).
Square-and-add: $R=\sqrt{3^2+4^2}=\sqrt{25}=5$ .
Divide: $\tan\alpha=\tfrac43\Rightarrow\alpha=53.13^\circ$ (acute).

Answer

3\sin\theta+4\cos\theta=5\sin(\theta+53.13^\circ)

Worked example

Real question: express $5\sin x-3\cos x$ as $R\sin(x-\alpha)$ , then state max and min

9709/31 O/N 2021 Q2. It leads with sin and the question wants a minus inside, so the target is $R\sin(x-\alpha)$ .

Expand the target: $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$ , so $R\cos\alpha=5$ and $R\sin\alpha=3$ .
$R=\sqrt{5^2+3^2}=\sqrt{34}$ (exact), and $\tan\alpha=\tfrac35\Rightarrow\alpha=30.96^\circ$ .
A sine runs from $-1$ to $+1$ , so $5\sin x-3\cos x$ runs from $-\sqrt{34}$ to $+\sqrt{34}$ .

Answer

\sqrt{34}\sin(x-30.96^\circ);\ \max=\sqrt{34},\ \min=-\sqrt{34}

Worked example

Solve $2\sin\theta-3\cos\theta=\sqrt2$ for $0^\circ<\theta<360^\circ$

Fold the left side (lead with sin, minus inside → $R\sin(\theta-\alpha)$ ): $R=\sqrt{13},\ \tan\alpha=\tfrac32\Rightarrow\alpha=56.31^\circ$ , so the equation becomes $\sqrt{13}\sin(\theta-56.31^\circ)=\sqrt2$ .
Isolate the sine: $\sin(\theta-56.31^\circ)=\sqrt{\tfrac{2}{13}}=0.3922$ .
Solve for the bracket: $\theta-56.31^\circ=23.09^\circ$ or $180^\circ-23.09^\circ=156.91^\circ$ .
Add $56.31^\circ$ back to each: $\theta=79.4^\circ$ or $213.2^\circ$ .

Answer

\theta=79.4^\circ\quad\text{or}\quad \theta=213.2^\circ

Worked example

Express $2\cos\theta-\sin\theta$ as $R\cos(\theta+\alpha)$ , then give its max and min

It leads with cos, so target $R\cos(\theta+\alpha)=R\cos\alpha\cos\theta-R\sin\alpha\sin\theta$ . Match: $R\cos\alpha=2$ , $R\sin\alpha=1$ .
$R=\sqrt{4+1}=\sqrt5$ , $\tan\alpha=\tfrac12\Rightarrow\alpha=26.57^\circ$ .
A cosine runs from $-1$ to $+1$ , so the max is $\sqrt5$ (at $\theta=333.4^\circ$ ) and the min is $-\sqrt5$ (at $\theta=153.4^\circ$ ).

Answer

\sqrt5\cos(\theta+26.57^\circ);\ \max=\sqrt5,\ \min=-\sqrt5

Why the max is R and where it sits

Reading the answer straight off the folded form

Where the marks go

Common mistake

Keep

\alpha

in the first quadrant (

0^\circ<\alpha<90^\circ

): take the acute

\tan^{-1}\tfrac{b}{a}

, not a second-quadrant value.

Common mistake

The

\sin\theta

coefficient is

R\cos\alpha

(not

R\sin\alpha

). Write the expansion out in full before matching, or the whole answer flips.

Common mistake

When the question says exact, leave

R

as a surd like

\sqrt{34}

: it is

\alpha

that you give to 2 dp.

Now you try

Show that $5\sec x+\tan x=4$ can be written as $R\cos(x+\alpha)=5$ , giving the exact $R$ and $\alpha$ to 2 dp $(0^\circ<\alpha<90^\circ)$ . (9709/33 Oct/Nov 2022 Q7)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2022 Q7)

Multiply through by $\cos x$ ( $\sec x=\tfrac1{\cos x}$ , $\tan x=\tfrac{\sin x}{\cos x}$ ): $5+\sin x=4\cos x$ , so $4\cos x-\sin x=5$ .

It starts with cos, so fold the left side: $R=\sqrt{4^2+1^2}=\sqrt{17},\ \tan\alpha=\tfrac14\Rightarrow\alpha=14.04^\circ$ :

Answer

\sqrt{17}\cos(x+14.04^\circ)=5