A LevelMathsCambridge (CIE)Pure 3FlashcardsTrigonometryCompound & Double Angle Formulae

Compound & Double Angle Formulae (Cambridge (CIE) A Level Maths: Pure 3): Flashcards

Exam code: 9709

1/19

0Still learning

Know0

FrontCompound Angle Formulae
$\sin (A + B) \equiv \sin A \cos B +______$
$\cos (A + B) \equiv \cos A \cos B -______$

Was this flashcard helpful?

Cards in this collection (19)

$\sin (A + B) \equiv \sin A \cos B +______$
$\cos (A + B) \equiv \cos A \cos B -______$
$\sin (A + B) \equiv \sin A \cos B + \cos A \sin B$
$\cos (A + B) \equiv \cos A \cos B - \sin A \sin B$
For $\sin$ , the sign on the right matches the one on the left.
How do the signs work in the $\tan$ compound angle formula?
The sign on the left matches the one in the numerator, and is opposite to the one in the denominator.
So $\tan (A + B) \equiv \frac{\tan A + \tan B}{1 - \tan A \tan B}$ .
How is the $\tan$ compound angle formula derived?
Write $\tan (A + B)$ as $\frac{\sin (A + B)}{\cos (A + B)}$ and expand both parts.
Dividing the numerator and denominator by $\cos A \cos B$ then turns every term into a tangent.
True or False?
$\cos (A - B) \equiv \cos A \cos B + \sin A \sin B$
True.
For cosine the sign on the right is opposite to the one on the left, so a minus outside gives a plus inside.
Sine and tangent behave the other way round, which is why the cosine pair is the one that catches people out.
Why is $\sin (A + B)$ not simply $\sin A + \sin B$ ?
Because sine is not additive: adding the angles does not add the values.
Arithmetic settles it at once. $\sin 90 ° = 1$ , but $\sin 45 ° + \sin 45 ° \approx 1.41$ .
How can a compound angle formula give the exact value of $\sin 75 \circ$ ?
Write $75 °$ as $45 ° + 30 °$ , both of which have exact values.
Expanding $\sin (45 ° + 30 °)$ then gives an exact answer in surds.
Complete the double angle formula for tangent:
$\tan 2 A \equiv \frac{______}{1 -______}$
The completed formula is:
$\tan 2 A \equiv \frac{2 \tan A}{1 - \tan^{2} A}$
Note that it is not simply $2 \tan A$ : doubling the angle does not double the tangent.
What are the three forms of $\cos 2 A$ ?
They are $\cos^{2} A - \sin^{2} A$ , $2 \cos^{2} A - 1$ and $1 - 2 \sin^{2} A$ .
The last two come from the first by using $\sin^{2} A + \cos^{2} A \equiv 1$ to remove one of the two squared terms.
How is every double angle formula obtained from a compound angle formula?
By setting $B = A$ , so that $A + B$ becomes $2 A$ .
Each of $\sin (A + A)$ , $\cos (A + A)$ and $\tan (A + A)$ then collapses into the corresponding double angle result.
True or False?
$\sin 4 A \equiv 2 \sin 2 A \cos 2 A$
True.
The formula holds for any angle, so replacing $A$ by $2 A$ throughout is perfectly legitimate.
The same move gives $\cos 6 A$ from the $\cos 2 A$ formula with $A$ replaced by $3 A$ .
Which form of $\cos 2 A$ should you choose?
Choose whichever form is written in terms of the function the rest of the equation already uses.
An equation containing $\sin A$ elsewhere is easiest with the form carrying only $\sin^{2} A$ , because everything is then in a single function.
What should you do on meeting $2 \sin A \cos A$ in an expression?
Replace it with $\sin 2 A$ , turning two factors into a single term.
The double angle formulae are just as useful read from right to left as from left to right.
What do you do with an equation containing both $\sin 2 θ$ and $\sin θ$ ?
Use a double angle formula on the $2 θ$ term, so that every term involves the same angle.
In $7 \sin 2 θ - 3 \sin θ = 0$ that gives $14 \sin θ \cos θ - 3 \sin θ = 0$ , which then factorises.
Define harmonic form.
Writing an expression such as $a \sin x + b \cos x$ as a single trigonometric function, $R \sin (x + α)$ .
It is the reverse of expanding with a compound angle formula, and can be thought of as factorising.
For $a \sin x + b \cos x \equiv R \sin (x + α)$ :
$R =______$ and $\tan α =______$
$R = \sqrt{a^{2} + b^{2}}$ and $\tan α = \frac{b}{a}$
$R$ is always taken positive, and $α$ between $0 °$ and $90 °$ .
How do you rewrite an expression in harmonic form?
Expand the target form with the appropriate compound angle formula, then equate coefficients of $\sin x$ and $\cos x$ .
That gives $R \cos α = a$ and $R \sin α = b$ , which between them fix both unknowns.
Why does dividing the two coefficient equations give you $\tan α$ ?
Because the $R$ s cancel, leaving $\frac{\sin α}{\cos α} = \frac{b}{a}$ .
That quotient is $\tan α$ , so $α = \tan^{- 1} \frac{b}{a}$ .
Why is harmonic form useful?
It turns two trigonometric terms into one, which can then be solved or analysed like any single function.
The maximum of $R \sin (x + α)$ is $R$ and its minimum is $- R$ , which is immediate once it is written that way.
True or False?
There is only one correct harmonic form for a given expression.
False.
The same expression can be written with $\sin$ or with $\cos$ , and with $+ α$ or $- α$ , giving four equivalent forms.
A question will say which of them it wants.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Reciprocal Trigonometric FunctionsNext:Further Trigonometric Equations