Applications of the Fundamental Identities (worked examples)

Applications of the Fundamental Identities (worked examples)

These examples show the standard moves used with $\sin^2\theta+\cos^2\theta=1$, $1+\tan^2\theta=\sec^2\theta$, $1+\cot^2\theta=\text{cosec}^2\theta$: squaring a given relation, forming a quadratic in one ratio, combining fractions over a common denominator, and eliminating θ between two parametric equations.

1. Given $\tan\theta+\dfrac1{\tan\theta}=2$, find $\tan^2\theta+\dfrac1{\tan^2\theta}$

Square both sides: $\tan^2\theta + 2\cdot\tan\theta\cdot\dfrac1{\tan\theta} + \dfrac1{\tan^2\theta} = 4$, i.e. $\tan^2\theta+2+\dfrac1{\tan^2\theta}=4$, so $\tan^2\theta+\dfrac1{\tan^2\theta}=2$.

2. Which of three proposed identities is actually true?

i) $2\cos^2\theta = \dfrac{1-\tan^2\theta}{1+\tan^2\theta}$? RHS $=\dfrac{1-\sin^2\theta/\cos^2\theta}{1+\sin^2\theta/\cos^2\theta} = \dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta} = \cos^2\theta-\sin^2\theta = \cos^2\theta-(1-\cos^2\theta)=2\cos^2\theta-1 \ne 2\cos^2\theta$ (LHS). Not true.

ii) $\dfrac{\cot A-\tan B}{\cot B-\tan A}=\cot A\tan B$? Substitute $A=B=45°$: LHS $=\dfrac{1-1}{1-1}=\dfrac00$, undefined/degenerate, while RHS $=\cot45°\tan45°=1$. Since one counter-example is enough to disprove a general claim, and the two sides disagree (the LHS doesn't even give a consistent value), this proposed identity is not true in general.

iii) $\dfrac{\cos\theta}{1-\tan\theta}+\dfrac{\sin\theta}{1-\cot\theta}=\sin\theta+\cos\theta$? Rewrite $1-\tan\theta = \dfrac{\cos\theta-\sin\theta}{\cos\theta}$ and $1-\cot\theta=\dfrac{\sin\theta-\cos\theta}{\sin\theta}$, so LHS $= \dfrac{\cos^2\theta}{\cos\theta-\sin\theta} + \dfrac{\sin^2\theta}{\sin\theta-\cos\theta} = \dfrac{\cos^2\theta-\sin^2\theta}{\cos\theta-\sin\theta} = \dfrac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{\cos\theta-\sin\theta} = \sin\theta+\cos\theta$ = RHS. True.

3. $5\tan A=\sqrt2$, $\pi<A<3\pi/2$; $\sec B=\sqrt{11}$, $3\pi/2<B<2\pi$; find $\text{cosec}\,A-\tan B$

$\tan A=\tfrac{\sqrt2}5$, so $\cot A=\tfrac5{\sqrt2}$ and $\text{cosec}^2A=1+\cot^2A=1+\tfrac{25}2=\tfrac{27}2$. Since A is in the third quadrant, cosecA is negative: $\text{cosec}\,A=-\sqrt{27/2}$. Also $\tan^2B=\sec^2B-1=11-1=10$; since B is in the fourth quadrant, $\tan B=-\sqrt{10}$. So $\text{cosec}\,A-\tan B = -\sqrt{27/2}-(-\sqrt{10}) = \sqrt{10}-\sqrt{27/2}$.

4. tanθ = 1/√7, evaluate $\dfrac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}$

Using $\text{cosec}^2\theta=1+\cot^2\theta$ and $\sec^2\theta=1+\tan^2\theta$: $\text{cosec}^2\theta-\sec^2\theta=\cot^2\theta-\tan^2\theta$ and $\text{cosec}^2\theta+\sec^2\theta=\cot^2\theta+\tan^2\theta+2$. With $\tan\theta=1/\sqrt7$, $\cot\theta=\sqrt7$: numerator $=7-1/7=48/7$; denominator $=7+1/7+2=64/7$. Ratio $=48/64=3/4$.

5. Prove $\cos^6\theta+\sin^6\theta = 1-3\sin^2\theta\cos^2\theta$

Using $a^3+b^3=(a+b)^3-3ab(a+b)$ with $a=\cos^2\theta,\ b=\sin^2\theta$: LHS $=(\cos^2\theta+\sin^2\theta)^3 - 3\cos^2\theta\sin^2\theta(\cos^2\theta+\sin^2\theta) = 1^3 - 3\sin^2\theta\cos^2\theta(1) = 1-3\sin^2\theta\cos^2\theta$ = RHS.

6. Eliminate θ

i) $x=a\cos\theta,\ y=b\sin\theta$: $\cos\theta=x/a,\ \sin\theta=y/b$; squaring and adding using $\sin^2\theta+\cos^2\theta=1$: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.

ii) $x=a\cos^3\theta,\ y=b\sin^3\theta$: $\cos\theta=(x/a)^{1/3},\ \sin\theta=(y/b)^{1/3}$; squaring and adding: $\left(\dfrac{x}{a}\right)^{2/3}+\left(\dfrac{y}{b}\right)^{2/3}=1$.

iii) $x=2+3\cos\theta,\ y=5+3\sin\theta$: $\cos\theta=\dfrac{x-2}3,\ \sin\theta=\dfrac{y-5}3$; squaring and adding: $\dfrac{(x-2)^2}{9}+\dfrac{(y-5)^2}9=1$, i.e. $(x-2)^2+(y-5)^2=9$.

7. $2\sin^2\theta+7\cos\theta=5$, find permissible cosθ

Substitute $\sin^2\theta=1-\cos^2\theta$: $2(1-\cos^2\theta)+7\cos\theta=5 \implies 2-2\cos^2\theta+7\cos\theta-5=0 \implies 2\cos^2\theta-7\cos\theta+3=0$. Factor: $2\cos^2\theta-6\cos\theta-\cos\theta+3=0 \implies (2\cos\theta-1)(\cos\theta-3)=0$, so $\cos\theta=3$ or $\cos\theta=\tfrac12$. Since $\cos\theta$ can never exceed 1, $\cos\theta=3$ is rejected. Permissible value: $\cos\theta=\tfrac12$.

8. Solve $4\sin^2\theta - 2(\sqrt3+1)\sin\theta+\sqrt3=0$ for θ

This is a quadratic in $\sin\theta$ with $a=4,\ b=-2(\sqrt3+1),\ c=\sqrt3$. By the quadratic formula, $\sin\theta = \dfrac{2(\sqrt3+1)\pm\sqrt{4(\sqrt3+1)^2-16\sqrt3}}{8}$. Since $(\sqrt3+1)^2=4+2\sqrt3$, the discriminant simplifies to $4(4+2\sqrt3)-16\sqrt3 = 16+8\sqrt3-16\sqrt3=16-8\sqrt3 = 4(\sqrt3-1)^2$, so $\sqrt{\text{disc}}=2(\sqrt3-1)$. Then $\sin\theta = \dfrac{2(\sqrt3+1)\pm2(\sqrt3-1)}{8} = \dfrac{(\sqrt3+1)\pm(\sqrt3-1)}{4}$, giving $\sin\theta=\dfrac{2\sqrt3}4=\dfrac{\sqrt3}2$ or $\sin\theta=\dfrac24=\dfrac12$. So $\theta=\dfrac{\pi}3$ or $\theta=\dfrac{\pi}6$.

9. $\tan\theta+\sec\theta=1.5$, find tanθ, sinθ, secθ

Write $\tan\theta+\sec\theta=\tfrac32$ … (1). Since $\sec^2\theta-\tan^2\theta=1$, $(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1$, so $\tfrac32(\sec\theta-\tan\theta)=1 \implies \sec\theta-\tan\theta=\tfrac23$ … (2). Adding (1) and (2): $2\sec\theta = \tfrac32+\tfrac23 = \tfrac{13}6 \implies \sec\theta=\tfrac{13}{12}$, so $\cos\theta=\tfrac{12}{13}$. Subtracting: $2\tan\theta=\tfrac32-\tfrac23=\tfrac56 \implies \tan\theta=\tfrac5{12}$, so $\sin\theta=\tan\theta\cos\theta = \tfrac5{12}\times\tfrac{12}{13}=\tfrac5{13}$.

10. Prove $\dfrac{\sin\theta}{1-\cos\theta}+\dfrac{\tan\theta}{1+\cos\theta} = \sec\theta\,\text{cosec}\,\theta+\cot\theta$ …