The six trigonometric ratios (right-triangle definition). For an acute angle θ in a right triangle, with sides labelled relative to θ as opposite, adjacent, and the hypotenuse:
Only sinθ and cosθ are truly independent — every other ratio is a quotient or reciprocal built from them, which is why rewriting a hard trig expression purely in sin,cos is such a reliable first move.
Exact values at standard angles (0∘,30∘,45∘,60∘,90∘):
θ
0∘
30∘
45∘
60∘
90∘
sinθ
0
21
21
23
1
cosθ
1
23
21
21
0
tanθ
0
31
1
3
undefined
tan90∘,sec90∘ are undefined because cos90∘=0; csc0∘,cot0∘ are undefined because sin0∘=0. Also sin30∘=cos60∘ and sin60∘=cos30∘ — an early instance of the general complementary-angle pattern sinθ=cos(90∘−θ).
What makes an equation an identity. A trigonometric identity is an equation in trigonometric ratios holding for every value of θ in its domain — not merely for some particular angle. secθ=cosθ1 is an identity (true for all θ with cosθ=0); sinθ=21 is not (true only at specific angles like 30∘ or 150∘).
The three fundamental (Pythagorean) identities, obtained from the Pythagorean theorem applied to a right triangle, divided in turn by the square of the hypotenuse, the adjacent side, and the opposite side:
cos2θ+sin2θ=1,sec2θ−tan2θ=1,csc2θ−cot2θ=1.
Here sin2θ means (sinθ)2, and similarly for the other ratios. Each identity holds wherever both sides are defined — e.g. sec2θ−tan2θ=1 says nothing at θ=90∘, where both terms are individually undefined, but this doesn't stop it from being a genuine identity for every θ where it does make sense.
Reciprocal and quotient identities (restating the ratio definitions as identities in their own right): …
Substituting x=sin2α,y=sin2β turns the given condition into a perfect square (x−y)2=0, forcing x=y; both required identities then reduce to trivial statements once sin2α=sin2β is known.
Step 1. Set up substitutions. Let x=sin2α (so cos2α=1−x) and y=sin2β (so cos2β=1−y). The given condition cos2βcos4α+sin2βsin4α=1 becomes 1−y(1−x)2+yx2=1.
Step 2. Clear denominators. Multiplying through by y(1−y):
(1−x)2y+x2(1−y)=y(1−y).
Expand the left side: y(1−2x+x2)+x2−x2y=y−2xy+x2y+x2−x2y=y−2xy+x2.
So the equation becomes y−2xy+x2=y−y2.
Step 3. Simplify. Cancel y from both sides: −2xy+x2=−y2⇒x2−2xy+y2=0⇒(x−y)2=0⇒x=y.
So the given condition forces sin2α=sin2β — and therefore also cos2α=1−x=1−y=cos2β.
Step 4. Prove (i). With sin2α=sin2β=x: sin4α+sin4β=x2+x2=2x2, and 2sin2αsin2β=2(x)(x)=2x2. These are equal, proving (i). …