Skip to content

Mathematics · Ch 2 — Inverse Trigonometric Functions

Elementary Properties and Identities

5

Elementary Properties and Identities

5. Elementary Properties and Identities

(A) Reciprocal identities. Since cosec θ=1sin⁡θ\text{cosec}\,\theta=\dfrac{1}{\sin\theta}, if x=cosec θx=\text{cosec}\,\theta then 1x=sin⁡θ\dfrac1x=\sin\theta, so

sin⁡−1 ⁣(1x)=cosec−1x,x≥1 or x≤−1,\sin^{-1}\!\left(\frac1x\right)=\text{cosec}^{-1}x,\qquad x\ge1\text{ or }x\le-1,

and by the identical argument,

cos⁡−1 ⁣(1x)=sec⁡−1x,x≥1 or x≤−1,tan⁡−1 ⁣(1x)=cot⁡−1x,x>0.\cos^{-1}\!\left(\frac1x\right)=\sec^{-1}x,\qquad x\ge1\text{ or }x\le-1,\qquad\qquad \tan^{-1}\!\left(\frac1x\right)=\cot^{-1}x,\qquad x>0.

(B) Negative-argument identities.

Odd-type (Proof for sin⁡−1\sin^{-1}). Let y=sin⁡−1xy=\sin^{-1}x, so x=sin⁡yx=\sin y with y∈[−π2,π2]y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. Then −x=−sin⁡y=sin⁡(−y)-x=-\sin y=\sin(-y), and since y∈[−π2,π2]y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right], so is −y-y. Hence −y-y is the principal value of sin⁡−1(−x)\sin^{-1}(-x), i.e.

sin⁡−1(−x)=−y=−sin⁡−1x,x∈[−1,1].■\sin^{-1}(-x)=-y=-\sin^{-1}x,\qquad x\in[-1,1].\qquad\blacksquare

The same argument (using that each branch is symmetric about the origin) gives

tan⁡−1(−x)=−tan⁡−1x  (x∈R),cosec−1(−x)=−cosec−1x  (∣x∣≥1).\tan^{-1}(-x)=-\tan^{-1}x\ \ (x\in\mathbb{R}),\qquad \text{cosec}^{-1}(-x)=-\text{cosec}^{-1}x\ \ (|x|\ge1).

π\pi-shift-type (Proof for cos⁡−1\cos^{-1}). Let y=cos⁡−1xy=\cos^{-1}x, so x=cos⁡yx=\cos y with y∈[0,π]y\in[0,\pi]. Then −x=−cos⁡y=cos⁡(π−y)-x=-\cos y=\cos(\pi-y), and since y∈[0,π]y\in[0,\pi], so is π−y\pi-y (as yy runs from 00 to π\pi, π−y\pi-y runs from π\pi to 00). Hence π−y\pi-y is the principal value of cos⁡−1(−x)\cos^{-1}(-x):

cos⁡−1(−x)=π−y=π−cos⁡−1x,x∈[−1,1].■\cos^{-1}(-x)=\pi-y=\pi-\cos^{-1}x,\qquad x\in[-1,1].\qquad\blacksquare

Because their branches [0,π][0,\pi] and (0,π)(0,\pi) are likewise not symmetric about the origin,

sec⁡−1(−x)=π−sec⁡−1x  (∣x∣≥1),cot⁡−1(−x)=π−cot⁡−1x  (x∈R).\sec^{-1}(-x)=\pi-\sec^{-1}x\ \ (|x|\ge1),\qquad \cot^{-1}(-x)=\pi-\cot^{-1}x\ \ (x\in\mathbb{R}).

(C) Complementary-angle sum identities.

Theorem. sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2} for every x∈[−1,1]x\in[-1,1].

Proof. Let y=sin⁡−1xy=\sin^{-1}x, so x=sin⁡yx=\sin y with y∈[−π2,π2]y\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]. By the co-function relation sin⁡y=cos⁡ ⁣(π2−y)\sin y=\cos\!\left(\dfrac{\pi}{2}-y\right),

x=cos⁡ ⁣(π2−y).x=\cos\!\left(\frac{\pi}{2}-y\right).

For this to identify π2−y\dfrac{\pi}{2}-y as the principal value of cos⁡−1x\cos^{-1}x, it must lie in [0,π][0,\pi]. Since y∈[−π2,π2]y\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right], multiplying by −1-1 and adding π2\dfrac{\pi}{2} gives π2−y∈[0,π]\dfrac{\pi}{2}-y\in[0,\pi] exactly (at y=−π2y=-\frac{\pi}{2}, π2−y=π\frac{\pi}{2}-y=\pi; at y=π2y=\frac{\pi}{2}, π2−y=0\frac{\pi}{2}-y=0). So π2−y\dfrac{\pi}{2}-y is the principal value of cos⁡−1x\cos^{-1}x:

cos⁡−1x=π2−y=π2−sin⁡−1x⟹sin⁡−1x+cos⁡−1x=π2.■\cos^{-1}x=\frac{\pi}{2}-y=\frac{\pi}{2}-\sin^{-1}x \quad\Longrightarrow\quad \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}.\qquad\blacksquare

The identical co-function argument (using tan⁡y=cot⁡(π2−y)\tan y=\cot\left(\frac{\pi}{2}-y\right) and sec⁡y=cosec(π2−y)\sec y=\text{cosec}\left(\frac{\pi}{2}-y\right), and checking the shifted angle lands in the right branch each time) proves the two companion identities:

tan⁡−1x+cot⁡−1x=π2  (x∈R),sec⁡−1x+cosec−1x=π2  (∣x∣≥1).\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\ \ (x\in\mathbb{R}),\qquad \sec^{-1}x+\text{cosec}^{-1}x=\frac{\pi}{2}\ \ (|x|\ge1).

(D) Addition formula for tan⁡−1\tan^{-1}. Let α=tan⁡−1x\alpha=\tan^{-1}x, β=tan⁡−1y\beta=\tan^{-1}y, so x=tan⁡αx=\tan\alpha, y=tan⁡βy=\tan\beta. By the tangent sum formula (Class XI),

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=x+y1−xy.\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{x+y}{1-xy}.

Provided xy<1xy<1 (which keeps α+β\alpha+\beta inside (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right), the principal branch of tan⁡−1\tan^{-1}), taking tan⁡−1\tan^{-1} of both sides is valid and gives

tan⁡−1x+tan⁡−1y=tan⁡−1 ⁣(x+y1−xy),xy<1.\tan^{-1}x+\tan^{-1}y=\tan^{-1}\!\left(\frac{x+y}{1-xy}\right),\qquad xy<1. …

Start your 14-day free trial to unlock the full solution →
← Previous · 4Graphs of Inverse Trigonometric FunctionsNext · summary →Summary