Hard A-Level Trigonometric graphs Questions

Recognise the hidden quadratic

The equation is a quadratic in $\cos x$. Writing $c=\cos x$ turns it into $2c^2+c-1=0$, which we can factorise like any quadratic.

Write the quadratic

Now it looks exactly like a standard quadratic equation in $c$.

Identify the coefficients

Comparing with $ac^2+bc+c_0=0$, the coefficients are $a=2$, $b=1$ and constant $-1$. We look for factors of $a\times(-1)=-2$ that add to $1$: these are $+2$ and $-1$.

Factorise

We look for two brackets that multiply to give $2c^2+c-1$. Checking: $(2c-1)(c+1)=2c^2+2c-c-1=2c^2+c-1$, correct.

Solve each bracket

Setting each factor to zero gives $\cos x=\tfrac12$ or $\cos x=-1$.

First value: find the acute angle

For $\cos x=\tfrac12$ the basic angle is $60^\circ$.

First value: quadrant solutions

Cosine is positive in the first and fourth quadrants, giving $60^\circ$ and $360^\circ-60^\circ=300^\circ$.

Second value: solve cosine = -1

Cosine reaches $-1$ only at the bottom of its cycle, at $x=180^\circ$.

Combine the solutions

Putting both cases together gives three solutions.

Check the factorisation

Expanding the brackets returns the original quadratic, so the factorisation was correct — a quick safeguard against slips.

Verify a solution

Substituting $x=60^\circ$ gives $\tfrac12+\tfrac12-1=0$, confirming this solution.

Verify the fourth-quadrant solution

Since $\cos 300^\circ=\tfrac12$ as well, substituting $x=300^\circ$ also gives $\tfrac12+\tfrac12-1=0$, confirming this solution.

Explain why three, not four

The value $\cos x=\tfrac12$ gives two angles, but $\cos x=-1$ gives only one (at $180^\circ$), so there are three solutions rather than four.

Interpret using the graph

The line $y=\tfrac12$ cuts the cosine graph twice and $y=-1$ just touches it at the bottom, giving three intersection points.

Check the range

All three values lie in the range.

List all solutions

So the solutions are $60^\circ$, $180^\circ$ and $300^\circ$.

Write tangent as a ratio

The key is that tangent is defined as sine divided by cosine. We test what happens when we add $180^\circ$ to $x$.

Shift sine by 180 degrees

Adding $180^\circ$ flips the sign of sine, because the sine graph is upside-down half a cycle later.

Shift cosine by 180 degrees

Adding $180^\circ$ also flips the sign of cosine, for the same reason.

Form the new tangent

Putting these into the ratio gives a negative over a negative.

Simplify

The two minus signs cancel, so $\tan(x+180^\circ)=\tan x$. The graph repeats after only $180^\circ$.

Check with a numerical example

Trying actual numbers, $\tan 30^\circ$ and $\tan(30^\circ+180^\circ)=\tan 210^\circ$ are both $\tfrac{1}{\sqrt3}$, confirming the repeat.

Show the period is not smaller

A shift of $90^\circ$ does not reproduce the graph (it gives $-\cot x$), so $180^\circ$ really is the smallest repeat.

Relate to the graph

On the graph, each branch between consecutive asymptotes looks exactly like the one $180^\circ$ earlier, which is what a period of $180^\circ$ means.

Reject the 'by definition' option

Saying the period equals the asymptote gap 'by definition' is circular — the gap is $180^\circ$ *because* of the sign cancellation, not the other way round.

Reject the 'signs unchanged' option

The option claiming $\sin(x+180^\circ)=\sin x$ is simply false; the sign does change, so that reasoning is wrong.

Reject the 'must be smaller' option

Having asymptotes does not by itself force a period smaller than $360^\circ$; the reason is the ratio of two sign-flipped functions.

Reject the 'every value once' option

Although tangent does take every value once per $180^\circ$, that is a restatement of the period, not the explanation for it.

Compare with sine and cosine

This is why tangent repeats twice as often as sine and cosine: their signs must both return to normal, which takes $360^\circ$, but the ratio recovers after only $180^\circ$.

Confirm it is the smallest repeat

Since no smaller positive shift reproduces the graph, $180^\circ$ is genuinely the period, not just a multiple of it.

State the conclusion

Because the value repeats every $180^\circ$, that is the period — half that of sine and cosine.

Start from the tangent graph

Ordinary $y=\tan x$ has asymptotes at $90^\circ$ and $270^\circ$ and passes through the origin. The bracket $x-45^\circ$ will shift it.

Interpret the shift

A subtraction inside the bracket moves the whole graph $45^\circ$ to the right.

Move the asymptotes

Each asymptote shifts right by $45^\circ$, so they are now at $135^\circ$ and $315^\circ$.

Find a key point

The curve crosses zero where the bracket is zero, i.e. at $x=45^\circ$ instead of $x=0^\circ$.

Check the value at 90 degrees

At $x=90^\circ$ the bracket is $45^\circ$, so $y=1$. The curve is rising as it approaches the first asymptote.

Check the value at 180 degrees

At $x=180^\circ$ the bracket is $135^\circ$, giving $y=-1$ on the branch just after the first asymptote.

Find the second zero

The curve crosses zero again at $x=225^\circ$, exactly $180^\circ$ (one tangent period) after the first zero.

Describe a branch

Between the asymptotes at $135^\circ$ and $315^\circ$ the curve climbs steadily from $-\infty$ up through $(225^\circ,0)$ to $+\infty$.

Tabulate key features

The whole graph is fixed by its zeros at $45^\circ$ and $225^\circ$ and its asymptotes at $135^\circ$ and $315^\circ$.

Reject the 'left shift' graph

A graph crossing zero at $x=315^\circ$ (shifted left) is $\tan(x+45^\circ)$, so it is wrong.

Reject the 'no shift' graph

A graph passing through the origin has not been shifted at all, so it is ordinary $\tan x$ — wrong.

Reject the reflected graph

A graph whose branches fall from left to right has also been reflected, so it is wrong.

Confirm the zeros

The correct graph passes through zero at $45^\circ$ and $225^\circ$, matching our working.

Confirm the asymptotes

It also has its vertical asymptotes at $135^\circ$ and $315^\circ$, exactly where the original ones moved to after the shift.

Select the matching graph

The correct graph is tangent shifted $45^\circ$ right, crossing zero at $45^\circ$ with asymptotes at $135^\circ$ and $315^\circ$.

Look for a link between sine and cosine

Sine and cosine are connected: the cosine of an angle equals the sine of its complement. Rewriting cosine as a sine lets us compare two sines directly.

Rewrite the right-hand side

Replacing $\cos x$ with $\sin(90^\circ-x)$ turns the equation into 'sine equals sine'.

Use the sine-equals-sine rule (case 1)

Two angles have the same sine if they are equal (up to full turns). This is the first case.

Solve case 1

Collecting the $x$ terms and dividing by 2 gives a family of solutions spaced $180^\circ$ apart.

Test the k values in range

Substituting $k=0,1,2$ into $x=15^\circ+180^\circ k$ gives $15^\circ$, $195^\circ$ and $375^\circ$; only the first two lie in $0^\circ\le x\le360^\circ$.

List case-1 solutions in range

So case 1 contributes $15^\circ$ and $195^\circ$; the value $375^\circ$ is rejected as out of range.

Use the sine-equals-sine rule (case 2)

The second case uses the supplement: the other angle is $180^\circ$ minus the first.

Simplify case 2

The $x$ terms cancel, leaving $60^\circ=90^\circ$, which is impossible.

Reject case 2

Since this statement is false, case 2 gives no solutions.

Verify the first solution

Checking $x=15^\circ$: $\sin(15^\circ+60^\circ)=\sin 75^\circ\approx0.966$ and $\cos 15^\circ\approx0.966$, which agree.

Verify the second solution

Checking $x=195^\circ$: $\sin(255^\circ)\approx-0.966$ and $\cos(195^\circ)\approx-0.966$, which also agree.

Interpret using the graphs

Drawing $y=\sin(x+60^\circ)$ and $y=\cos x$ on the same axes, they intersect exactly twice in $0^\circ$ to $360^\circ$, matching our two solutions.

Note the solution spacing

The two solutions differ by $180^\circ$, which is expected from the family $x=15^\circ+180^\circ k$ we found in case 1.

Count the solutions

Case 1 gives two solutions in range and case 2 gives none, so there are exactly two solutions altogether.

List all solutions

So the only solutions are $15^\circ$ and $195^\circ$.

Write the equation to solve

We want the height $h$ to equal 14, so we set the model equal to 14. This mixes a real context with trig-equation skills.

Move the constant

Subtract 10 from both sides to start isolating the cosine term.

Divide by -8

Dividing both sides by $-8$ gives $\cos\theta=-\tfrac12$. Take care with the sign when dividing by a negative number.

Recall the exact value

The size $\tfrac12$ is a standard exact value: $\cos 60^\circ=\tfrac12$, which fixes the related acute angle.

Find the acute angle

Ignoring the sign, the related acute angle is $60^\circ$.

Decide the quadrants

Because the value is negative, the solutions lie in the second and third quadrants.

Second-quadrant solution

This gives $\theta=120^\circ$.

Third-quadrant solution

This gives $\theta=240^\circ$.

Verify the first angle

Substituting $\theta=120^\circ$ gives $10+4=14$, exactly the height required.

Verify the second angle

Substituting $\theta=240^\circ$ also gives 14, confirming this solution.

Find the range of heights

Since $\cos\theta$ runs between $-1$ and 1, the height runs between $10-8=2$ m and $10+8=18$ m. A height of 14 m is within this, so solutions exist.

Explain the two answers physically

The passenger passes 14 m twice per turn: once on the way up and once on the way down. That is why there are two values of $\theta$.

Check the range

Both values lie in the allowed range for one turn of the wheel.

Note the symmetry

The two solutions are symmetric about $180^\circ$, since $240^\circ=360^\circ-120^\circ$ — a useful check that matches the shape of the cosine curve.

Interpret the answer

So the passenger is 14 m up when the wheel has turned $120^\circ$ and again at $240^\circ$ — once going up, once coming down.