Q.Show that $\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}=\tan^{-1}\frac{63}{16}$.

Concept and Intuition

Mixed inverse-trig equations become tractable once every inverse function is replaced by an angle with a concrete right triangle, so all the sines/cosines needed are just ratios of sides — no further inverse-trig identities are needed beyond $\sin(\alpha-\beta)$.

Step-by-Step Solution

1. The equation is $\tan^{-1}\!\left(\dfrac{1}{2\sqrt2}\right)-\cos^{-1}\!\left(\dfrac{1}{\sqrt3}\right)+\sin^{-1}x=0$, i.e. $\sin^{-1}x = \cos^{-1}\!\left(\dfrac1{\sqrt3}\right)-\tan^{-1}\!\left(\dfrac1{2\sqrt2}\right)$.
2. Let $\alpha=\cos^{-1}(1/\sqrt3)$: then $\cos\alpha=1/\sqrt3$, and $\sin\alpha=\sqrt{1-1/3}=\sqrt{2/3}=\sqrt2/\sqrt3$.
3. Let $\beta=\tan^{-1}(1/(2\sqrt2))$: right triangle with opposite $=1$, adjacent $=2\sqrt2$, hypotenuse $=\sqrt{1+8}=3$. So $\sin\beta=1/3$, $\cos\beta=2\sqrt2/3$.
4. $x=\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$. …

Concept and Intuition

For $\cos^{-1}A+\cos^{-1}B=C$, taking the cosine of both sides using $\cos(P+Q)=\cos P\cos Q-\sin P\sin Q$ (with $\sin(\cos^{-1}A)=\sqrt{1-A^2}$) converts the inverse-trig equation into an ordinary algebraic one.

Step-by-Step Solution

1. Let $A=\cos^{-1}2x,\ B=\cos^{-1}3x$, so $A+B=\pi/3$ and $\cos(A+B)=1/2$.
2. $\cos(A+B) = \cos A\cos B - \sin A\sin B = (2x)(3x) - \sqrt{1-4x^2}\sqrt{1-9x^2} = 6x^2 - \sqrt{(1-4x^2)(1-9x^2)}$.
3. So $6x^2 - \dfrac12 = \sqrt{(1-4x^2)(1-9x^2)}$.
4. Square both sides: $(6x^2-\tfrac12)^2 = (1-4x^2)(1-9x^2)$.
5. LHS $= 36x^4 - 6x^2 + \tfrac14$. RHS $= 1 - 13x^2 + 36x^4$.
6. Equating: $36x^4-6x^2+\tfrac14 = 36x^4-13x^2+1 \Rightarrow -6x^2+\tfrac14 = -13x^2+1 \Rightarrow 7x^2 = \tfrac34$. …

Concept and Intuition

Powers of $\cos$ at supplementary-type angles ($5\pi/8=\pi-3\pi/8$, $7\pi/8=\pi-\pi/8$) are related by a sign flip, which vanishes under an even power — this is what collapses four terms into two equal pairs. Then the double-angle identity $\cos^4\theta+\sin^4\theta=1-\tfrac12\sin^22\theta$ finishes the algebra.

Step-by-Step Solution

1. $\cos\frac{5\pi}8=\cos\left(\pi-\frac{3\pi}8\right)=-\cos\frac{3\pi}8$, and $\cos\frac{7\pi}8=-\cos\frac\pi8$. Raised to the 4th power, signs disappear:

2. $\cos\frac{3\pi}8=\sin\frac\pi8$, so $k=2\left(\cos^4\tfrac\pi8+\sin^4\tfrac\pi8\right)$.
3. Use $\cos^4\theta+\sin^4\theta = (\cos^2\theta+\sin^2\theta)^2-2\sin^2\theta\cos^2\theta = 1-\tfrac12\sin^22\theta$. With $\theta=\pi/8$: $\sin(\pi/4)=\frac{\sqrt2}{2}$, $\sin^2(\pi/4)=\tfrac12$, so this is $1-\tfrac14=\tfrac34$. …

Concept and Intuition

The expression inside $\cos^{-1}$ looks like $\dfrac12\sin\theta+\dfrac{\sqrt3}2\cos\theta$ once we set $x=\sin\theta$ — a classic "$R\sin(\theta+\phi)$" combination with $R=1$, $\phi=60^\circ$. Recognizing this collapses the whole messy expression to a single clean angle.

Step-by-Step Solution

1. Let $\theta=\sin^{-1}x$, so $x=\sin\theta$ and (since $0<x<1/2\Rightarrow 0<\theta<\pi/6$) $\cos\theta=\sqrt{1-x^2}>0$.
2. $\sqrt{3-3x^2}=\sqrt3\sqrt{1-x^2}=\sqrt3\cos\theta$, so the argument of $\cos^{-1}$ is $\dfrac x2+\dfrac{\sqrt3\cos\theta}2=\dfrac12\sin\theta+\dfrac{\sqrt3}2\cos\theta$.
3. Recognize this as $\sin\theta\cos60^\circ+\cos\theta\sin60^\circ=\sin(\theta+60^\circ)=\sin\left(\theta+\dfrac\pi3\right)$.
4. Convert to a cosine so $\cos^{-1}$ can undo it directly: $\sin\left(\theta+\dfrac\pi3\right)=\cos\left(\dfrac\pi2-\theta-\dfrac\pi3\right)=\cos\left(\dfrac\pi6-\theta\right)$. …

Concept and Intuition

Just like circular functions, hyperbolic functions have an addition formula. Given $\sinh a$ or $\cosh b$, the companion function follows from the hyperbolic Pythagorean identity $\cosh^2-\sinh^2=1$.

Step-by-Step Solution

1. $a=\sinh^{-1}\sqrt8 \Rightarrow \sinh a=\sqrt8$. Then $\cosh a=\sqrt{1+\sinh^2a}=\sqrt{1+8}=\sqrt9=3$.
2. $b=\cosh^{-1}5 \Rightarrow \cosh b=5$. Then $\sinh b=\sqrt{\cosh^2b-1}=\sqrt{25-1}=\sqrt{24}=2\sqrt6$.
3. Addition formula: $\cosh(a+b)=\cosh a\cosh b+\sinh a\sinh b = (3)(5)+(\sqrt8)(2\sqrt6)$.
4. $(3)(5)=15$. $(\sqrt8)(2\sqrt6)=2\sqrt{48}=2\cdot4\sqrt3=8\sqrt3$. …

Concept and Intuition

Just as with inverse trig functions, when you're given $\sinh^{-1}$ of specific numbers and asked about the hyperbolic cosine of their sum, the cleanest route is the hyperbolic angle-addition identity $\cosh(A+B)=\cosh A\cosh B+\sinh A\sinh B$ — you never need to write out the logarithmic form of $\sinh^{-1}$.

Step-by-Step Solution

1. Let $A=\sinh^{-1}(2)$, so $\sinh A=2$. Using $\cosh^2A-\sinh^2A=1$: $\cosh A=\sqrt{1+4}=\sqrt5$ (positive since $\cosh\ge1$ always).
2. Let $B=\sinh^{-1}(3)$, so $\sinh B=3$, and $\cosh B=\sqrt{1+9}=\sqrt{10}$.
3. Given $\alpha=A+B$, apply the addition formula: $\cosh\alpha=\cosh(A+B)=\cosh A\cosh B+\sinh A\sinh B$. …

Concept and Intuition

The identity $\cos^{-1}(2x^2-1)=2\cos^{-1}x$ only holds when $x\ge0$ (so that $2\theta$ stays within $[0,\pi]$, the range of $\cos^{-1}$). When $x<0$, $\theta=\cos^{-1}x$ lies in $(\frac\pi2,\pi)$, so $2\theta\in(\pi,2\pi)$ falls outside the principal range, and we must instead use $\cos^{-1}(\cos2\theta) = 2\pi-2\theta$ for that range.

Step-by-Step Solution

1. Since $n\in N$ (so $n\ge1$), $x=-\dfrac{n}{n+1}$ is always negative and lies in $[-\frac12, -1)$... more precisely in $(-1,0)$, approaching $-1$ as $n\to\infty$ and equal to $-\frac12$ at $n=1$.
2. Let $\theta=\cos^{-1}x$. Since $x<0$, $\theta\in\left(\dfrac\pi2,\pi\right)$.
3. $\cos2\theta = 2\cos^2\theta-1 = 2x^2-1$.
4. Since $\theta\in(\frac\pi2,\pi)$, $2\theta\in(\pi,2\pi)$ — outside $[0,\pi]$, the range of the principal $\cos^{-1}$.
5. To bring $2\theta$ back to $[0,\pi]$: since $\cos(2\pi-2\theta)=\cos2\theta$ and $2\pi-2\theta\in(0,\pi)$ (as $2\theta\in(\pi,2\pi)$), we get $\cos^{-1}(2x^2-1) = \cos^{-1}(\cos2\theta) = 2\pi-2\theta$.
6. So $2\cos^{-1}x + \cos^{-1}(2x^2-1) = 2\theta + (2\pi-2\theta) = 2\pi$. …

Concept and Intuition

Inverse-trig equations need two checks: first the domain (arguments must lie in $[-1,1]$), then verification that any algebraic solution obtained by taking cosines/squaring actually satisfies the ORIGINAL equation (squaring can introduce extraneous roots, and taking cosine of both sides of $\cos^{-1}(\cdot)=(\text{expr})$ only recovers a necessary condition).

Step-by-Step Solution

1. Domain: need $1-x\in[-1,1]\Rightarrow x\in[0,2]$ and $x\in[-1,1]$ for $\cos^{-1}x$. Combined: $x\in[0,1]$.
2. Let $\theta=\cos^{-1}x$, so $\theta\in[0,\pi/2]$ for $x\in[0,1]$. The equation becomes $\cos^{-1}(1-x)=\tfrac{\pi}{2}+2\theta$.
3. Take cosine of both sides: $1-x=\cos\left(\tfrac{\pi}{2}+2\theta\right)=-\sin2\theta=-2\sin\theta\cos\theta=-2x\sqrt{1-x^2}$ (since $\cos\theta=x$, $\sin\theta=\sqrt{1-x^2}\ge0$).
4. Squaring: $(1-x)^2=4x^2(1-x^2)=4x^2(1-x)(1+x)$. Factor: $(1-x)\left[(1-x)-4x^2(1+x)\right]=0$, giving $x=1$ or $4x^3+4x^2+x-1=0$. …