trigonometric-functions se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
658 questions tagged with trigonometric-functions.
What is \(\frac{1-\cos 2x}{\sin 2x}\) equal to?
D. \(\tan x\)
\(1-\cos 2x=2\sin^2 x\) और \(\sin 2x=2\sin x\cos x\) लगाएं। सरलीकरण से \(\tan x\) मिलता है। / Use \(1-\cos 2x=2\sin^2 x\) and \(\sin 2x=2\sin x\cos x\). Simplification gives \(\tan x\).
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What is \(\frac{\sin 2x}{1+\cos 2x}\) equal to?
C. \(\tan x\)
\(\sin 2x=2\sin x\cos x\) और \(1+\cos 2x=2\cos^2 x\) रखें। परिणाम \(\tan x\) है। / Use \(\sin 2x=2\sin x\cos x\) and \(1+\cos 2x=2\cos^2 x\). The result is \(\tan x\).
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If \(\sec x+\tan x=3\), what is the value of \(\sec x-\tan x\)?
B. \( \frac{1}{3} \)
क्योंकि (\(\sec x+\tan x\)\(\sec x-\tan x\)=1), इसलिए दूसरा गुणक \(\frac{1}{3}\) है। परीक्षा में reciprocal pair पहचानें। / Since (\(\sec x+\tan x\)\(\sec x-\tan x\)=1), the other factor is \(\frac{1}{3}\). In exams, identify reciprocal pairs.
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What is the fundamental period of (f(x)=\sin-2 x)?
A. \( \pi \)
क्योंकि (\sin-2\(x+\pi\)=\sin-2 x), इसलिए मूल आवर्त \(\pi\) है। परीक्षा में वर्ग वाले sine और cosine का आवर्त याद रखें। / Since (\sin-2\(x+\pi\)=\sin-2 x), the fundamental period is \(\pi\). In exams, remember the period of squared sine and cosine.
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What is the fundamental period of (f(x)=|\sin x|)?
D. \( \pi \)
\(|\sin x|\) में ऋणात्मक भाग धनात्मक हो जाता है, इसलिए पैटर्न हर \(\pi\) पर दोहरता है। परीक्षा में modulus से आवर्त घट सकता है। / In \(|\sin x|\), the negative part becomes positive, so the pattern repeats every \(\pi\). In exams, modulus can reduce the period.
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What is the product form of \(\cos 4x-\cos 2x\)?
B. \(-2\sin 3x\sin x\)
सूत्र \(\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\) है। इसलिए उत्तर \(-2\sin 3x\sin x\) है। / The formula is \(\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\). Hence the answer is \(-2\sin 3x\sin x\).
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What is the product form of \(\sin x+\sin 3x\)?
A. \(2\sin 2x\cos x\)
सूत्र \(\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\) लगाएं। यहाँ उत्तर \(2\sin 2x\cos x\) है। / Use \(\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\). Here the answer is \(2\sin 2x\cos x\).
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If \(\tan x=2\), what is the value of \(\frac{1-\tan^2 x}{1+\tan^2 x}\)?
A. \( -\frac{3}{5} \)
सीधा रखने पर \(\frac{1-4}{1+4}=-\frac{3}{5}\) मिलता है। यह \(\cos 2x\) का भी रूप है। / Direct substitution gives \(\frac{1-4}{1+4}=-\frac{3}{5}\). This is also the form of \(\cos 2x\).
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What is the correct simplified form of (\cos\(\pi+x\))?
D. \(-\cos x\)
\(\pi+x\) तीसरे चतुर्थांश से जुड़ा रूप है जहाँ \(\cos x\) ऋणात्मक होता है। इसलिए (\cos\(\pi+x\)=-\cos x)। / The form \(\pi+x\) is linked to the third quadrant where \(\cos x\) is negative. Hence (\cos\(\pi+x\)=-\cos x).
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What is (\sin\left\(\frac{\pi}{2}+x\right\)) equal to?
C. \(\cos x\)
\(\frac{\pi}{2}+x\) के लिए sine, cosine में बदलता है और दूसरे चतुर्थांश में धनात्मक रहता है। इसलिए मान \(\cos x\) है। / For \(\frac{\pi}{2}+x\), sine changes to cosine and remains positive in the second quadrant. Hence the value is \(\cos x\).
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If \(\sin x=\frac{3}{5}\) and (x) lies in the first quadrant, what is the value of \(\tan x\)?
A. \( \frac{3}{4} \)
\(\cos x=\frac{4}{5}\) होगा, इसलिए \(\tan x=\frac{\sin x}{\cos x}=\frac{3}{4}\)। परीक्षा में चतुर्थांश से चिन्ह तय करें। / We get \(\cos x=\frac{4}{5}\), so \(\tan x=\frac{\sin x}{\cos x}=\frac{3}{4}\). In exams, decide the sign from the quadrant.
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What type of function is (f(x)=\sin x\cos x)?
B. विषम फलनOdd function
(f(-x)=\sin(-x)\cos(-x)=-\sin x\cos x=-f(x)) है। इसलिए यह विषम फलन है। / Here (f(-x)=\sin(-x)\cos(-x)=-\sin x\cos x=-f(x)). So it is an odd function.
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What is the range of (f(x)=\sin x+\cos x)?
D. \( [-\sqrt{2},\sqrt{2}] \)
(\sin x+\cos x=\sqrt{2}\sin\left\(x+\frac{\pi}{4}\right\)) होता है। इसलिए परिसर \( [-\sqrt{2},\sqrt{2}] \) है। / We have (\sin x+\cos x=\sqrt{2}\sin\left\(x+\frac{\pi}{4}\right\)). Hence the range is \( [-\sqrt{2},\sqrt{2}] \).
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What is the fundamental period of (f(x)=\tan\frac{x}{2}+\cot\frac{x}{2})?
C. \( 2\pi \)
\(\tan\frac{x}{2}\) और \(\cot\frac{x}{2}\) दोनों का आवर्त \(2\pi\) है। परीक्षा में \(\tan ax\) का आवर्त \( \frac{\pi}{|a|} \) लगाएं। / Both \(\tan\frac{x}{2}\) and \(\cot\frac{x}{2}\) have period \(2\pi\). In exams, use the period of \(\tan ax\) as \( \frac{\pi}{|a|} \).
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What is the value of \(\sin^2 20^\circ+\sin^2 70^\circ\)?
B. (1)
क्योंकि \(\sin 70^\circ=\cos 20^\circ\), इसलिए योग \(\sin^2 20^\circ+\cos^2 20^\circ=1\) है। परीक्षा में complementary angles पहचानें। / Since \(\sin 70^\circ=\cos 20^\circ\), the sum is \(\sin^2 20^\circ+\cos^2 20^\circ=1\). In exams, identify complementary angles.
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If \(\tan \theta+\cot \theta=4\), what is the value of \(\tan^2\theta+\cot^2\theta\)?
A. (14)
(\(\tan \theta+\cot \theta\)2=\tan-2\theta+\cot-2\theta+2) होता है। इसलिए मान (16-2=14) है। / We use (\(\tan \theta+\cot \theta\)2=\tan-2\theta+\cot-2\theta+2). Hence the value is (16-2=14).
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What is the minimum value of (f(x)=4+3\cos 2x)?
D. (1)
\(\cos 2x\) का न्यूनतम मान (-1) है, इसलिए (4+3(-1)=1)। परीक्षा में अधिकतम और न्यूनतम के लिए \(\cos x\) की सीमा याद रखें। / The minimum value of \(\cos 2x\) is (-1), so (4+3(-1)=1). In exams, remember the range of \(\cos x\) for maximum and minimum.
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How many solutions does \(\sin x=\cos x\) have in the interval \(0\leq x<2\pi\)?
C. (2)
\(\sin x=\cos x\) से \(\tan x=1\) मिलता है, इसलिए \(x=\frac{\pi}{4},\frac{5\pi}{4}\) हैं। परीक्षा में दिए गए अंतराल को ध्यान से देखें। / From \(\sin x=\cos x\), we get \(\tan x=1\), so \(x=\frac{\pi}{4},\frac{5\pi}{4}\). In exams, check the given interval carefully.
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What is the range of the function (f(x)=3+2\sin x)?
B. ( [1,5] )
क्योंकि \(-1\leq \sin x \leq 1\), इसलिए \(3+2\sin x\) का मान (1) से (5) तक होगा। परीक्षा में पहले मूल फलन का परिसर लिखें। / Since \(-1\leq \sin x \leq 1\), \(3+2\sin x\) ranges from (1) to (5). In exams, write the base range first.
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If (f(x)=\sin 3x+\cos 5x), what is the fundamental period of (f(x))?
C. \( 2\pi \)
\(\sin 3x\) और \(\cos 5x\) के आवर्त क्रमशः \( \frac{2\pi}{3} \) और \( \frac{2\pi}{5} \) हैं, जिनका सामान्य आवर्त \(2\pi\) है। परीक्षा में गुणांकों के लिए आवर्त अलग-अलग निकालें। / The periods of \(\sin 3x\) and \(\cos 5x\) are \( \frac{2\pi}{3} \) and \( \frac{2\pi}{5} \), whose common period is \(2\pi\). In exams, find each period separately first.
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What is (\tan\(5\pi+\theta\)) equal to?
C. \(\tan \theta\)
\(\tan \theta\) का मूल period \(\pi\) है और \(5\pi\) period का गुणज है, इसलिए मान नहीं बदलता। परीक्षा में tangent में \(\pi\) के गुणज को हटाकर सरल करें। / The fundamental period of \(\tan \theta\) is \(\pi\), and \(5\pi\) is a multiple of the period, so the value does not change. In exams simplify tangent by removing multiples of \(\pi\).
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A. \(\sec \theta\) धनात्मक और \(\cosec \theta\) ऋणात्मक\(\sec \theta\) positive and \(\cosec \theta\) negative
चतुर्थ चतुर्थांश में \(\cos \theta\) धनात्मक और \(\sin \theta\) ऋणात्मक होता है, इसलिए उनके व्युत्क्रमों के चिन्ह भी वैसे ही होंगे। परीक्षा में पहले मूल फलनों के चिन्ह तय करें। / In the fourth quadrant, \(\cos \theta\) is positive and \(\sin \theta\) is negative, so their reciprocals have the same signs. In exams decide the signs of basic functions first.
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What is the fundamental period of (\cos\(4\theta\))?
A. \(\frac{\pi}{2}\)
\(\cos k\theta\) का मूल period \(\frac{2\pi}{k}\) होता है, इसलिए यहां \(\frac{\pi}{2}\) मिलेगा। परीक्षा में coefficient को denominator में रखें। / The fundamental period of \( \cos k\theta\) is \( \frac{2\pi}{k}\), so here it is \( \frac{\pi}{2}\). In exams put the coefficient in the denominator.
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If \(\cos \theta=-1\), what is the general solution for \(\theta\)?
B. (\theta=(2n+1)\pi)
\(\cos \theta=-1\) विषम गुणजों of \(\pi\) पर होता है। परीक्षा में \(\cos \theta=1\) और \(\cos \theta=-1\) के हल अलग पहचानें। / \( \cos \theta=-1\) occurs at odd multiples of \( \pi\). In exams distinguish the solutions of \( \cos \theta=1\) and \( \cos \theta=-1\).
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What is the fundamental period of \(\sin 3\theta\)?
A. \(\frac{2\pi}{3}\)
\(\sin k\theta\) का period \(\frac{2\pi}{k}\) होता है, इसलिए यहां \(\frac{2\pi}{3}\) मिलेगा। परीक्षा में coefficient को period formula में लगाएं। / The period of \( \sin k\theta\) is \( \frac{2\pi}{k}\), so here it is \( \frac{2\pi}{3}\). In exams use the coefficient in the period formula.
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If \(\theta\) is in the third quadrant, what are the signs of \(\sin \theta\) and \(\cos \theta\)?
B. \(\sin \theta\) ऋणात्मक और \(\cos \theta\) ऋणात्मक\(\sin \theta\) negative and \(\cos \theta\) negative
तृतीय चतुर्थांश में sine और cosine दोनों ऋणात्मक होते हैं। परीक्षा में ASTC नियम से signs जल्दी तय करें। / In the third quadrant, both sine and cosine are negative. In exams use the ASTC rule to decide signs quickly.
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If the minimum value of \(5\cos \theta-12\sin \theta\) is (m), what is (m)?
B. (-13)
Amplitude (\sqrt{52+(-12)2}=13) है, इसलिए न्यूनतम (-13) होगा। परीक्षा में maximum positive और minimum negative amplitude होता है। / The amplitude is ( \sqrt{52+(-12)2}=13), so the minimum is (-13). In exams maximum is positive amplitude and minimum is negative amplitude.
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If the maximum value of \(3\sin \theta+4\cos \theta\) is (M), what is (M)?
A. (5)
\(a\sin \theta+b\cos \theta\) का अधिकतम \(\sqrt{a^2+b^2}\) होता है, इसलिए (M=5)। परीक्षा में coefficients का square जोड़ें। / The maximum of \(a\sin \theta+b\cos \theta\) is \( \sqrt{a^2+b^2}\), so (M=5). In exams add the squares of coefficients.
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What is the minimum value of \(\sin \theta-\cos \theta\)?
A. \(-\sqrt{2}\)
\(\sin \theta-\cos \theta\) का amplitude (\sqrt{12+(-1)2}=\sqrt{2}) है। परीक्षा में minimum हमेशा negative amplitude होगा। / The amplitude of \( \sin \theta-\cos \theta\) is ( \sqrt{12+(-1)2}=\sqrt{2}). In exams the minimum is the negative amplitude.
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What is the maximum value of \(\sin \theta+\cos \theta\)?
B. \(\sqrt{2}\)
(\sin \theta+\cos \theta=\sqrt{2}\sin\left\(\theta+\frac{\pi}{4}\right\)), इसलिए अधिकतम \(\sqrt{2}\) है। परीक्षा में \(a\sin x+b\cos x\) का maximum \(\sqrt{a^2+b^2}\) लें। / ( \sin \theta+\cos \theta=\sqrt{2}\sin\left\(\theta+\frac{\pi}{4}\right\)), so the maximum is \( \sqrt{2}\). In exams use maximum of \(a\sin x+b\cos x\) as \( \sqrt{a^2+b^2}\).
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