Exercises — Pythagorean identities — sin² + cos² = 1, derivations of the other two

WHAT we want: the missing squared term. WHY: it is literally the master identity. ${\sin }^2\theta +{\cos }^2\theta =1\Rightarrow \text{blank}={\cos }^2\theta .$

Start from the "tan–sec house" identity $1+{\tan }^2\theta ={\sec }^2\theta$. Subtract ${\tan }^2\theta$ from both sides: ${\sec }^2\theta -{\tan }^2\theta =1.$

Subtract ${\cos }^2\theta$ from both sides: ${\sin }^2\theta =1-{\cos }^2\theta .$ This is the form you will reach for whenever a problem hands you $\cos \theta$ and asks about $\sin \theta$.

WHAT: isolate ${\cos }^2\theta$. WHY: the master identity links the two. ${\cos }^2\theta =1-{\sin }^2\theta =1-\frac{25}{169}=\frac{144}{169}.$ Take the square root — but which sign? $\theta$ acute means Q1, and on the sign chart Q1 has $\cos >0$: $\cos \theta =+\frac{12}{13}.$

WHY use the cot–csc house? Because it connects exactly the two functions in play: $1+{\cot }^2\theta ={\csc }^2\theta .$ ${\csc }^2\theta =1+\frac{9}{16}=\frac{25}{16}\Rightarrow \csc \theta =+\frac{5}{4}(\text{Q1}\Rightarrow \sin >0\Rightarrow \csc >0).$ Since $\csc \theta =\frac{1}{\sin \theta }$, flip it: $\sin \theta =\frac{4}{5}$.

Use $1+{\tan }^2\theta ={\sec }^2\theta$ (it directly ties $\tan$ to $\sec =1/\cos$): ${\sec }^2\theta =1+\frac{49}{576}=\frac{625}{576}\Rightarrow \sec \theta =+\frac{25}{24}.$ Flip: $\cos \theta =\frac{1}{\sec \theta }=\frac{24}{25}$ (positive, as Q1 demands).

${\cos }^2\theta =1-\frac{9}{25}=\frac{16}{25}\Rightarrow \cos \theta =\pm \frac{4}{5}.$ WHICH sign? Q2 has $x<0$, so $\cos <0$: $\cos \theta =-\frac{4}{5}.$ Then $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{3/5}{-4/5}=-\frac{3}{4}.$ (Sanity: in Q2, $\sin >0$ but $\cos <0$, so their ratio $\tan$ is negative. ✓)

${\sin }^2\theta =1-\frac{144}{169}=\frac{25}{169}\Rightarrow \sin \theta =\pm \frac{5}{13}.$ Q3 has $y<0$, so $\sin <0$: $\sin \theta =-\frac{5}{13},\csc \theta =\frac{1}{\sin \theta }=-\frac{13}{5}.$

${\sec }^2\theta =1+{\tan }^2\theta =1+\frac{9}{16}=\frac{25}{16}\Rightarrow \sec \theta =\pm \frac{5}{4}.$ Q4 has $x>0$, so $\cos >0$, hence $\sec >0$: $\sec \theta =+\frac{5}{4}$, giving $\cos \theta =\frac{4}{5}$. Now $\sin \theta =\tan \theta \cdot \cos \theta =\left(-\frac{3}{4}\right)\left(\frac{4}{5}\right)=-\frac{3}{5}$. (Check Q4: $\cos >0$, $\sin <0$ ✓; and ${\sin }^2+{\cos }^2=\frac{9}{25}+\frac{16}{25}=1$ ✓.)

WHAT we do: rewrite each bracket using a Pythagorean identity.

• $1-{\sin }^2\theta ={\cos }^2\theta$ (master identity rearranged).
• $1+{\tan }^2\theta ={\sec }^2\theta$ (tan–sec house).

So the product is ${\cos }^2\theta \cdot {\sec }^2\theta ={\cos }^2\theta \cdot \frac{1}{{\cos }^2\theta }=1.■$ WHY it works: $\sec$ is the reciprocal of $\cos$, so the two ${\cos }^2$ factors cancel exactly.

Strategy: cross-multiplying is risky (could introduce false roots), so instead multiply the left side top and bottom by the conjugate $1+\sin \theta$: $\frac{\cos \theta }{1-\sin \theta }\cdot \frac{1+\sin \theta }{1+\sin \theta }=\frac{\cos \theta (1+\sin \theta )}{1-{\sin }^2\theta }.$ WHY the conjugate? It turns the bottom into a difference of squares $1-{\sin }^2\theta$, which the master identity converts to ${\cos }^2\theta$: $=\frac{\cos \theta (1+\sin \theta )}{{\cos }^2\theta }=\frac{1+\sin \theta }{\cos \theta }.■$

Factor the left side by pulling out ${\sec }^2\theta$: ${\sec }^2\theta ({\sec }^2\theta -1).$ Now use ${\sec }^2\theta -1={\tan }^2\theta$ (tan–sec house rearranged), and replace the leading ${\sec }^2\theta$ with $1+{\tan }^2\theta$: $(1+{\tan }^2\theta ){\tan }^2\theta ={\tan }^2\theta +{\tan }^4\theta ={\tan }^4\theta +{\tan }^2\theta .■$

Idea: square the given sum, because squaring creates a ${\sin }^2+{\cos }^2$ we can replace by $1$. $(\sin \theta +\cos \theta {)}^2={\sin }^2\theta +2\sin \theta \cos \theta +{\cos }^2\theta =\frac{1}{25}.$ Replace ${\sin }^2\theta +{\cos }^2\theta =1$: $1+2\sin \theta \cos \theta =\frac{1}{25}\Rightarrow 2\sin \theta \cos \theta =-\frac{24}{25}\Rightarrow \sin \theta \cos \theta =-\frac{12}{25}.$

Common denominator (a difference of squares appears): $\frac{(1-\sin \theta )+(1+\sin \theta )}{(1+\sin \theta )(1-\sin \theta )}=\frac{2}{1-{\sin }^2\theta }.$ Use $1-{\sin }^2\theta ={\cos }^2\theta$: $=\frac{2}{{\cos }^2\theta }=2{\sec }^2\theta .$

First part. At $\theta =90°$ the point on the circle is $(0,1)$, so $\cos 90°=0,\sin 90°=1$. Then ${\sin }^{2}90°+{\cos }^{2}90°=1^2+0^2=1.✓$ The master identity survives every angle, including this degenerate one, because it only uses $\sin$ and $\cos$, which are defined everywhere. Second part. $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and $\sec \theta =\frac{1}{\cos \theta }$. At $\theta =90°$, $\cos \theta =0$, so both $\tan$ and $\sec$ divide by zero — they are undefined. An identity can only be applied where every function in it exists, so $1+{\tan }^2\theta ={\sec }^2\theta$ has no meaning at $90°$. (This is why the tan–sec house always carries the condition $\cos \theta \ne 0$.)

$\csc \theta =\frac{1}{\sin \theta }$, so $\sin \theta =\frac{8}{17}$ (positive — consistent with Q2, where $\sin >0$). ${\cos }^2\theta =1-\frac{64}{289}=\frac{225}{289}\Rightarrow \cos \theta =\pm \frac{15}{17}.$ Q2 has $\cos <0$: $\cos \theta =-\frac{15}{17}$. Then $\cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{-15/17}{8/17}=-\frac{15}{8}.$