Let’s remind ourselves of the three key formulae that we use to solve to solve trigonometric equations:
Solve 4sin5x – 2 = 0
Solve 2cos5x + 1 = 0
Short Recap Exercise
Introduction of Reciprocal Trigonometric Functions
Each trigonometric function has a reciprocal:
We can divide Pythagoras’ Theorem by sin2x or cos2x to give us alternative forms of Pythagoras’ Theorem that involve these reciprocal functions.
We can derive the graphs of the reciprocal trigonometric functions from those of the principal functions.
- Find cosec 120° as an exact number;
- Find values of in the interval for which .
Compound Angle Formulae
We need to memorise the below compound angle formulae, which we will use extensively. They can be derived geometrically using the rules from the right angled triangle, however this geometric derivation will not be covered during this course.
- sin(A-B)=sinAcosB – cosAsinB
- cos(A-B)=cosAcosB + sinAsinB
Express the following as single sine, cosine or tangent functions:
- sin15°cos20° + cos15°sin20°
- sin58°cos23° – cos58°sin23°
- cos130°cos80° – sin130°sin80°
Without using a calculator, find the exact value of:
We should also remember the following:
So if an angle is acute, then the sine, cosine and tangent of that angle will all be positive.
Likewise, if an angle is obtuse, then only the sine of the angle will be positive, it’s tangent and cosine will both be negative.
Given that 2sin(x+y)=3cos(x-y), express tanx in terms of tany
Double Angle Formulae
We can easily derive the double angle formulae for cos2x, sin2x and tan2x from the compound angle formulae.
We can then use Pythagoras’ Theorem to find alternative versions of cos 2x.
Each of the above formulae should be memorised as although we can derive it, we also need to be able to spot it in the context of a question.
Note: Take care not to divide an equation through by a trigonometric function, as this can lose some of the solutions to that equation. Instead, rearrange the equation using addition and subtraction and factorise out the trigonometric function, then these solutions won’t be lost.
Use the double-angle formulae to write each of the following as single trigonometric ratios:
- cos250° – sin250°
- Given that and , eliminate and express y in terms of x
- Given that cosx=3/4, and that 180° < x < 360°, find the exact value of sin2x and of tan 2x.
Using Trigonometric Identities to prove other Formulae
When asked to prove an equation, we take one side of the equation and change it to equivalent forms until eventually it is the same as the other side of the equation.
One common tactic is to replace tanx with sinx/cosx. Another is to use Pythagoras’ Theorem.
Care is needed when using the cosine double angle formulae because there are three alternatives.
Prove the identity
Prove the identity
Answers (all proofs so no formal specified answer)
Expressing asinx+bcosx in the form Rsin(x+𝛼)
By expanding out Rsin(x+𝛼) and then comparing it with asinx+bcosx, we can identify the values of R and 𝛼 and so rewrite this addition as a single function.
We can similarly use Rsin(x-𝛼), Rcos(x+𝛼) and Rcos(x-𝛼) to deal with asinx-bcosx, acosx-bsinx and acosx+bsinx.
This form can also help us find the minimum and the maximum values of a function (i.e. -R and R)
- Show that you can express 3sinx + 4 cosx in the form:
- Rcos(x-𝛼), where R>0 and 0°<𝛼<90°, giving your values of R and 𝛼 to 1 decimal place where appropriate.
- Show that you can express sinx – √3cosx in the form Rsin(x-𝛼), where R>0 and 0<𝛼<π/2. Hence sketch the graph of y = sinx – √3cosx
- Express 2cos𝜽 + 5 sin𝜽 in the form Rcos(𝜽-𝛼), where R>0 and 0°<𝛼<90°
- Hence solve for 0°<𝜽< 360° the equation 2cos𝜽 + 5sin𝜽 = 3
- f(𝜽) = 12cos𝜽 + 5sin𝜽
- Write f(𝜽) in the forrm Rcos(𝜽-𝛼)
- Find the maximum value of f(𝜽) and the smallest positive value of 𝜽 at which it occurs.
Answers (Miscellaneous Exercise)