Porisma

Use compound angles and R-formulae

Retain angle signs, derive equivalent forms and use amplitude to bound an expression.

Retain the addition and double-angle formulae

Choose the formula that reduces the number of different trigonometric functions. To reduce an even power, use a cosine double-angle form and repeat if needed.

The source labels both formula groups “in MF27”. The paired signs in cosine and tangent switch as indicated by ∓:

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B,cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B,tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B.\begin{aligned} \sin(A\pm B)&=\sin A\cos B\pm\cos A\sin B,\\ \cos(A\pm B)&=\cos A\cos B\mp\sin A\sin B,\\ \tan(A\pm B)&=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}. \end{aligned}

The double-angle formulae follow by taking equal angles:

sin⁡2A=2sin⁡Acos⁡A,cos⁡2A=cos⁡2A−sin⁡2A=1−2sin⁡2A=2cos⁡2A−1,tan⁡2A=2tan⁡A1−tan⁡2A.\begin{aligned} \sin2A&=2\sin A\cos A,\\ \cos2A&=\cos^2A-\sin^2A=1-2\sin^2A=2\cos^2A-1,\\ \tan2A&=\frac{2\tan A}{1-\tan^2A}. \end{aligned}

Tangent expressions require all terms and denominators involved to be defined. Example 20 retains the full triple-angle proof, correcting the source’s “sine terms only” callout to cosine terms. Example 21 retains both double-angle rearrangements that reduce cos⁴θ. Continue with the distinct Tutorial Question 14, which instead reduces sin⁴(θ/2).

Derive all four R-formulae

Expand the proposed shifted function and match its sine and cosine coefficients. Find the positive amplitude first, then the phase in the correct angle unit.

For positive a,b, write acos⁡x+bsin⁡x=Rcos⁡(x−α)a\cos x+b\sin x=R\cos(x-\alpha) with R>0R>0 and acute α. Expanding gives

acos⁡x+bsin⁡x=R(cos⁡xcos⁡α+sin⁡xsin⁡α)=Rcos⁡αcos⁡x+Rsin⁡αsin⁡x.\begin{aligned} a\cos x+b\sin x&=R(\cos x\cos\alpha+\sin x\sin\alpha)\\&=R\cos\alpha\cos x+R\sin\alpha\sin x. \end{aligned}

Comparing coefficients gives (1) Rcos⁡α=aR\cos\alpha=a and (2) Rsin⁡α=bR\sin\alpha=b. Divide (2) by (1) to obtain tan⁡α=b/a\tan\alpha=b/a, hence α=tan⁡−1(b/a)\alpha=\tan^{-1}(b/a). Adding their squares gives

R2cos⁡2α+R2sin⁡2α=a2+b2 ⟹ R2=a2+b2 ⟹ R=a2+b2.R^2\cos^2\alpha+R^2\sin^2\alpha=a^2+b^2 \ \Longrightarrow\ R^2=a^2+b^2 \ \Longrightarrow\ R=\sqrt{a^2+b^2}.

All four printed variants use this same R and α:

acos⁡x+bsin⁡x=Rcos⁡(x−α),asin⁡x+bcos⁡x=Rsin⁡(x+α),asin⁡x−bcos⁡x=Rsin⁡(x−α),acos⁡x−bsin⁡x=Rcos⁡(x+α).\begin{aligned} a\cos x+b\sin x&=R\cos(x-\alpha),\\ a\sin x+b\cos x&=R\sin(x+\alpha),\\ a\sin x-b\cos x&=R\sin(x-\alpha),\\ a\cos x-b\sin x&=R\cos(x+\alpha). \end{aligned}

Examples 22–23 preserve both extrema and the full shifted-bound chain. Their printed 56.3° and 22.6° phases are approximations; retain exact arctangents for identities, and use the same unit for the input angle and its phase. Continue with the distinct Tutorial Question 15.

Try these examples

Open an example in chapter practice. Try it first, then read its solution. Use “Learn this step” to return to this unit.