# Trigonometric Identities and formulas

There are so many identities like $\sin2θ$, $\cos2θ$, $\tan2θ$, $\sin(θ/2)$, $\cos(θ/2)$ and $\tan(θ/2)$. there are other formulas too like $\cos(α-β)$, $\sin(α-β)$ etc and yes the sum and product formulas of trigonometric function...

I am stuck in memorizing all. Is there any simple trick to memorize these formulas and identities?

• Practice is one way to go about it! :) – MonK Jul 7 '14 at 9:24

For the multiple angle stuff it suffices to remember De Moivre's formula: $$(\cos x + i \sin x)^n = \cos (nx) + i \sin (nx).\,$$ With it you easily get \begin{align}(\cos x + i \sin x)^2 &= \cos^2 x + 2i\sin x \cos x - \sin^2 x = (\cos^2 x - \sin^2 x) + i(2 \sin x \cos x)\\ &= \cos(2x) + i \sin (2x)\end{align} and then $$\cos^2 x - \sin^2 x = \cos(2x) \\ 2\sin x \cos x = \sin(2x) .$$ by comparing real and imaginary parts...

Sums and differences can be written in Matrix Form: \begin{align} & {} \quad \left(\begin{array}{rr} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{array}\right) \left(\begin{array}{rr} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{array}\right) \\[12pt] & = \left(\begin{array}{rr} \cos\alpha\cos\beta - \sin\alpha\sin\beta & -\cos\alpha\sin\beta - \sin\alpha\cos\beta \\ \sin\alpha\cos\beta + \cos\alpha\sin\beta & -\sin\alpha\sin\beta + \cos\alpha\cos\beta \end{array}\right) \\[12pt] & = \left(\begin{array}{rr} \cos(\alpha+\beta) & -\sin(\alpha+\beta) \\ \sin(\alpha+\beta) & \cos(\alpha+\beta) \end{array}\right). \end{align}

• @ draks : ur method is exact and accurate. Better to memorize Euler's formula. – user142971 Jul 7 '14 at 9:03
• @user36790 thx but it's not mine, it's ours... – draks ... Jul 7 '14 at 9:04
• @user159627 you're welcome... – draks ... Jul 7 '14 at 12:36

Prove all the formula by urself using concept of geometry especially the formula of compound angles. When to solve problems,first take help by directly seeing the formula. Don't cram them. Before sleeping try to memorize them . [Some techniques : $sin2\theta$ & $cos2\theta$ when expressed in terms of $tan\theta$ ,then in denominator, there will be always $+$ sign but in formula of $tan2\theta$, the denominator carries $-$ sign (tan is a hospitable guy!) . When $cos2\theta$ is expressed in terms of $sin^2\theta$ then sine takes $-$ sign but when expressed in terms of $cos^2\theta$ ,then cosine takes $+$ formula( it is cos' formula ; why should it be negative ?) . In formula of $sin3\theta$ it takes $3sin\theta$ as its 1st term followed by $4sin^3\theta$ . On the other hand, $cos3\theta$ takes $4cos^3\theta$ followed by $3cos\theta$ ( Cos is greedy; it takes the term beginning with 4 and then takes the term beginning with 3. But sin is sober ; it takes first the term beginning with 3 and then the other term).] .

• @ user 159627: Like Sid, the most simple & effective way to memorize them is to practise. Solve easy questions firstly. Remember problems of trigonometry can be solved effectively only by the proper application of proper formula at proper place . For instance $cos2\theta$ has 3 different forms; it will take long time(yes,u can solve it though) if u use improper form of $cos2\theta$ . But not to fear bcuz 'practice makes a man perfect' ! – user142971 Jul 7 '14 at 9:50
• @ 159627 : Best wishes for u. If any problem arises, just post it and i will help u! – user142971 Jul 8 '14 at 5:06

Here is how to get the compound angle formulas (including double/half/sum/difference……).

[Since you are only looking for a way to help you to memorize the result, the development involved will not be backed by reasons.]

Among all these formulas, only p: sin (A + B) = sin A cos B + cos A sin B needs to be memorized and all the others are just its “derivatives”.

Putting B = –b in (p) to get p1: sin (A – b)...............(*)

Differentiating (p) [wrt to A only] to get q : cos (A + B)

Same as (*) to get r : cos(A – B)

s: tan (A + B) is just (p) / (q). And tan (A – B) can be obtained similarly.

By letting A = B in (p), (q) and (s) to obtain the corresponding double angle formulas. In particular, $t_1: cos 2A = 1 – 2 sin^2 A$ and $t_2: cos 2A = 2 cos^2 A – 1$.

By re-arranging terms in $t_1$ and $t_2$, formulas for $sin^2 A$ and $cos^2 A$ can then be obtained.

Performing (p) + (p1), we get the product-to-(sum & difference) formula. Other formulas can be obtained similarly.

Sum-to-product can also be done with a little imagination. Hope you can take it from here.

• yeah. Thank you Mick. – user159627 Jul 7 '14 at 12:39
• You are welcome. Hope it helps. – Mick Jul 7 '14 at 14:35