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Transformation Formulae · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 19 questions.

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Question 12 Marks
Prove that:
$\sin65^\circ+\cos65^\circ=\sqrt2\cos20​​^\circ$
Answer
We have,
$\text{LHS}=\sin 65^\circ+\cos65^\circ$
$=\ \sin(45^\circ+20^\circ)+\cos(90^\circ-25^\circ)$  
$=\ \sin(45^\circ+20^\circ)+\sin25^\circ$
$=\ \sin(45^\circ+20^\circ)+\sin(45^\circ-20^\circ)$
$=\ 2\sin45^\circ\cos20^\circ$
$=\ 2\times\frac{1}{\sqrt2}\cos20^\circ$
$=\ \frac{\sqrt2\times\sqrt2}{\sqrt2}\times\cos20^\circ$
$=\ \sqrt2\cos20^\circ$
$=\ \text{RHS}$
$\therefore\ \sin60^\circ+\cos65^\circ=\sqrt2\cos20^\circ\ \text{Hence proved}.$
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Question 22 Marks
Prove that:
$\sin80^\circ-\cos70^\circ=\cos50^\circ$
Answer
$\sin80^\circ-\cos70^\circ=\cos50^\circ$
$\text{LHS}=\sin80^\circ=\cos50^\circ+\cos70^\circ$
Now,
$\cos\text{C}+\cos\text{D}=2\cos\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}$
$\text{RHS}=\cos50^\circ+\cos70^\circ$
$=2\cos\Big(\frac{50^\circ+70^\circ}{2}\Big)\cos\Big(\frac{50^\circ-70^\circ}{2}\Big)$
$=\ 2\cos60^\circ\cos(-10^\circ)$
$=\ 2\times\frac{1}{2}\cos10^\circ$ $[\cos(-\theta)=\cos\theta]$
$=\ \cos10^\circ$
$=\ \sin80^\circ$
$=\ \text{LHS}$ $[\because\ \cos\theta=\sin(90-\theta)]$
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Question 32 Marks
Prove that:
$\sin47^\circ+\cos77^\circ=\cos17^\circ$
Answer
We have,
$\text{LHS}=\sin47^\circ+\cos77^\circ$
$=\ \sin(90^\circ-43^\circ)+\cos77^\circ$
$=\ \cos43^\circ+\cos77^\circ$
$=\ \cos(60^\circ-17^\circ)+\cos(60^\circ+17^\circ)$
$=\ 2\cos60^\circ\cos17^\circ$
$=\ 2\times\frac{1}{2}\times\cos17^\circ$
$=\ \cos17^\circ$
$=\ \text{RHS}$
$\therefore\ \sin47^\circ+\cos77^\circ=\cos17^\circ\ \text{Hence proved}.$
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Question 42 Marks
Prove that:
$\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}=\sqrt3\sin\frac{\pi}{9}$
Answer
$\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}=\sqrt3\sin\frac{\pi}{9}$
$\text{LHS}=\sin\frac{5\pi}{18}-\cos\frac{4\pi}{9}$
$=\ \sin50^\circ-\cos80^\circ$
$=\ \sin50^\circ-\sin10^\circ$
$=\ 2\sin\Big(\frac{50^\circ-10^\circ}{2}\Big)\cos\Big(\frac{50^\circ+10^\circ}{2}\Big)$
$=\ 2\sin20^\circ\cos30^\circ$
$=\ 2\sin20^\circ\times\frac{\sqrt3}{2}$
$=\ \sqrt3\sin\frac{\pi}{9}=\text{RHS}$
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Question 52 Marks
Prove that:
$\sin105^\circ+\cos105^\circ=\cos45^\circ$
Answer
$\text{LHS}=\sin105^\circ+\cos105^\circ$
$\sin105^\circ+\cos(90^\circ+15^\circ)$
$\sin105^\circ-\sin15^\circ$
$=\ 2\sin\Big(\frac{105^\circ-15^\circ}{2}\Big)\cos\Big(\frac{105^\circ+15^\circ}{2}\Big)$
$=\ 2\sin45^\circ\cos60^\circ$
$=\ 2\frac{1}{\sqrt2}\frac{1}{2}$
$=\ \frac{1}{\sqrt2}$
$=\ \cos45^\circ$
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Question 62 Marks
Prove that:
$\cos80^\circ+\cos40^\circ-\cos20^\circ=0$
Answer
$\text{LHS}=\cos80^\circ+\cos40^\circ-\cos20^\circ$
$(\cos80^\circ+\cos40^\circ)-\cos20^\circ$
$=\ 2\cos\Big(\frac{80^\circ+40^\circ}{2}\Big)\cos\Big(\frac{80^\circ-40^\circ}{2}\Big)-\cos20^\circ$ $\Big[\because\ \sin\text{c}-\sin\text{D}=2\sin\Big(\frac{\text{C}-\text{D}}{2}\Big)\cos\Big(\frac{\text{C+D}}{2}\Big)\Big]$
$=\ 2\cos60^\circ\cos20^\circ-\cos20^\circ$
$=\ 2\times\frac{1}{2}\cos20-\cos20^\circ$
$=\ \cos20^\circ-\cos20^\circ$ 
$=\ 0$
$=\ \text{RHS}$
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Question 72 Marks
Prove that:
$\cos100^\circ+\cos20^\circ=\cos40^\circ$
Answer
$\cos100^\circ+\cos20^\circ=\cos40^\circ$
$\text{LHS}=\cos100^\circ+\cos20^\circ$ $[\because\ \cos\text{C}+\cos\text{D}=2\cos\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}]$
$\Rightarrow\ 2\cos\frac{(100^\circ+20^\circ)}{2}\cos\frac{100^\circ-20^\circ}{2}$
$=\ 2\cos60^\circ\cos40^\circ$
$=\ 2\times\frac{1}{2}\cos40^\circ$ $\Big[\because\ \cos60^\circ=\frac{1}{2}\Big]$
$=\ \cos40^\circ=\text{RHS}$
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Question 82 Marks
Prove that:
$\sin38^\circ+\sin22^\circ=\sin82^\circ$
Answer
$\sin38^\circ+\sin22^\circ=\sin82^\circ$
$\text{LHS}=\sin38^\circ+\sin22^\circ$
$\because\ \sin\text{C}\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C-D}}{2}$
$\Rightarrow\ \sin38^\circ+\sin22^\circ=2\sin\frac{60^\circ}{2}\cos\frac{16^\circ}{2}$
$=\ 2\sin30^\circ\cos8^\circ$
$=\ 2\times\frac{1}{2}\cos8^\circ$
$=\ \cos(90-82)^\circ$
$=\ \sin82^\circ=\text{RHS}$ $[\because\ \cos\text{x}=\sin(90-\text{x})]$
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Question 92 Marks
Show that:
$\sin50^\circ\cos85^\circ=\frac{1-\sqrt2\sin35^\circ}{2\sqrt2}$
Answer
$\sin50^\circ\cos85^\circ=\frac{1-\sqrt2\sin35^\circ}{2\sqrt2}$
$\text{LHS}=\sin50^\circ\cos85^\circ=\frac{2\sin50^\circ\cos85^\circ}{2}$
$\because\ 2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$
$\Rightarrow\ \frac{2\sin50^\circ\cos85^\circ}{2}=\frac{1}{2}[\sin(50^\circ+85^\circ)+\sin(50^\circ-85^\circ)]$
$=\ \frac{1}{2}[\sin135^\circ+\sin(-35^\circ)]$
$=\ \frac{1}{2}[\sin(90^\circ+45^\circ)-\sin35^\circ]$$[\because\ \sin(-\theta)=-\sin\theta]$
$=\ \frac{1}{2}[\cos45^\circ-\sin35^\circ]$$[\because\ \sin(90^\circ+\theta)=\cos\theta]$
Now,
$\cos45^\circ=\frac{1}{\sqrt2}$
$=\ \frac{1}{2}\Big[\frac{1}{\sqrt2}-\sin35^\circ\Big]$
$=\ \frac{1-\sqrt2\sin35^\circ}{2\sqrt2}=\text{RHS}$
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Question 102 Marks
Prove that:
$\sin23^\circ+\sin37^\circ=\cos7^\circ$
Answer
$\sin23^\circ+\sin37^\circ=\cos7^\circ$
$\text{LHS}=\sin23^\circ+\sin37^\circ$
$=\ 2\sin\Big(\frac{23^\circ+37^\circ}{2}\Big)\cos\Big(\frac{23^\circ-37^\circ}{2}\Big)$ $\Big[\because\ \sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}\Big]$
$=\ 2\sin(30^\circ)\cos(-7^\circ)$
$=\ 2\times\frac{1}{2}\cos7^\circ\Big[\because\ \cos(-\theta)=\cos\theta,\sin30^\circ=\frac{1}{2}\Big]$
$=\ \cos7^\circ=\text{RHS}$
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Question 112 Marks
Prove that:
$\sin50^\circ-\sin70^\circ+\sin10^\circ=0$
Answer
$\text{LHS}=\sin50^\circ-\sin70^\circ+\sin10^\circ$
$(\sin50^\circ-\sin70^\circ)+\sin10^\circ$
$=\ \Big(2\sin\Big(\frac{50^\circ-70^\circ}{2}\Big)\cos\Big(\frac{50^\circ+70^\circ}{2}\Big)\Big)+\sin10^\circ$ $\Big[\because\ \sin\text{c}-\sin\text{D}=2\sin\Big(\frac{\text{C}-\text{D}}{2}\Big)\cos\Big(\frac{\text{C+D}}{2}\Big)\Big]$
$=\ 2\sin(-10^\circ)\cos60^\circ+\sin10^\circ$
$=\ -2\sin10^\circ\times\frac{1}{2}+\sin10^\circ$ $\Big[\because\ \cos60^\circ=\frac{1}{2}\Big]$
$=\ 0$
$=\ \text{RHS}$
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Question 122 Marks
Prove that:
$\sin51^\circ+\cos81^\circ=\cos21^\circ$
Answer
Consider LHS:
$\sin51^\circ+\cos81^\circ$
$=\ \sin51^\circ+\cos(90^\circ-9^\circ)$
$=\ \sin51^\circ+\sin9^\circ$
$=\ 2\sin\Big(\frac{51^\circ+9^\circ}{2}\Big)\cos\Big(\frac{51^\circ-9^\circ}{2}\Big)$ $\Big\{\because\ \sin\text{A}+\sin\text{B}=-2\sin\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big\}$
$=\ 2\sin30^\circ\cos21^\circ$
$=\ 2\times\frac{1}{2}\cos(21^\circ)$
$=\ \sin(21^\circ)$
$=\ \text{RHS}$
Hence, LHS = RHS.
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Question 132 Marks
Prove that:
$\cos\Big(\frac{\pi}{4}+\text{x}\Big)-\cos\Big(\frac{\pi}{4}-\text{x}\Big)=\sqrt2\cos\text{x}$
Answer
We have,
$\text{LHS}=\cos\Big(\frac{\pi}{4}-\text{x}\Big)+\cos\Big(\frac{\pi}{4}+\text{x}\Big)$
$=\ 2\cos\frac{\pi}{4}\cos\text{x}$ $[\because\ \cos(\text{A}+\text{B})+\cos(\text{A}-\text{B})=2\cos\text{A}\cos\text{B}]$ 
$=\ 2\times\frac{1}{\sqrt2}\times\cos\text{x}$
$=\ \frac{\sqrt2\times\sqrt2}{\sqrt2}\cos\text{x}$
$=\ \sqrt2\cos\text{x}$
$=\ \text{RHS}$
$\therefore\ \cos\Big(\frac{\pi}{4}+\text{x}\Big)+\cos\Big(\frac{\pi}{4}-\text{x}\Big)=\sqrt2\cos\text{x}.$
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Question 142 Marks
Prove that:
$\cos55^\circ+\cos65^\circ\cos175^\circ=0$
Answer
Consider LHS:
$\cos55^\circ+\cos65^\circ+\cos175^\circ$
$=\ 2\cos\Big(\frac{55^\circ+65^\circ}{2}\Big)\cos\Big(\frac{55^\circ-65^\circ}{2}\Big)+\cos175^\circ$ $\Big\{\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big\}$
$=\ 2\cos60^\circ\cos(-5^\circ)+\cos175^\circ$
$=\ 2\times\frac{1}{2}\cos5^\circ+\cos175^\circ$
$=\ \cos5^\circ+\cos175^\circ$
$=\ 2\cos\Big(\frac{5^\circ+175^\circ}{2}\Big)\cos\Big(\frac{5^\circ-175^\circ}{2}\Big)$
$=\ 2\cos90^\circ\cos85^\circ$
$=\ 0=\text{RHS}$
Hence, LHS = RHS.
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Question 152 Marks
Prove that:
$\sin40^\circ+\sin20^\circ=\cos10^\circ$
Answer
$\sin40^\circ+\sin20^\circ=\cos10^\circ$
$\text{LHS}=\sin40^\circ+\sin20^\circ$
$=\ 2\sin\Big(\frac{40^\circ+20^\circ}{2}\Big)\cos\Big(\frac{40^\circ-20^\circ}{2}\Big)$ $\Big[\because\ \sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}\Big]$
$=\ 2\sin30^\circ\cos10^\circ$
$=\ 2\times\frac{1}{2}\cos10^\circ$
$=\ \cos10^\circ$
$=\ \text{RHS}$ $\Big[\because\ \sin30^\circ=\frac{1}{2}\Big]$
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Question 162 Marks
Prove that:
$\sin50^\circ+\sin10^\circ=\cos20^\circ$
Answer
$\sin50^\circ+\sin10^\circ=\cos20^\circ$
$\text{LHS}=\sin50^\circ+\sin10^\circ\ \Big[\because\ \sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}\Big]$
$\sin50^\circ+\sin10^\circ=2\sin\frac{60^\circ}{2}\cos20^\circ$
$=\ 2\sin30^\circ\cos20^\circ$
$=\ 2\times\frac{1}{2}\cos20^\circ$
$=\ \cos20^\circ=\text{RHS}\ \Big[\because30^\circ=\frac{1}{2}\Big]$
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Question 172 Marks
Prove that:
$\cos20^\circ+\cos100^\circ+\cos140^\circ=0$
Answer
$\text{LHS}=\cos20^\circ+\cos100^\circ+\cos140^\circ$
$\Rightarrow\ (\cos20^\circ+\cos100^\circ)+\cos140^\circ$
$=\ 2\cos\Big(\frac{20^\circ+100^\circ}{2}\Big)\cos\Big(\frac{20^\circ-100^\circ}{2}\Big)+\cos140^\circ$ $\Big[\because\ \cos\text{c}-\cos\text{D}=2\cos\Big(\frac{\text{C}-\text{D}}{2}\Big)\cos\Big(\frac{\text{C}-\text{D}}{2}\Big)\Big]$
$=\ 2\cos60^\circ\cos(-40^\circ)+\cos140^\circ$
$=\ 2\times\frac{1}{2}\cos40+\cos140^\circ$ $\Big[\because\ \cos60^\circ=\frac{1}{2}\Big]$
$=\ \cos40^\circ+\cos(180^\circ-40^\circ)$
$=\ \cos40^\circ-\cos40^\circ$ 
$=\ 0$
$=\ \text{RHS}$
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Question 182 Marks
Prove that:
$\cos\Big(\frac{3\pi}{4}+\text{x}\Big)-\cos\Big(\frac{3\pi}{4}-\text{x}\Big)=-\sqrt2\sin\text{x}$
Answer
We have,
$\text{LHS}=\cos\Big(\frac{3\pi}{4}-\text{x}\Big)-\cos\Big(\frac{3\pi}{4}+\text{x}\Big)$
$=\ -\Big[\cos\Big(\frac{3\pi}{4}-\text{x}\Big)-\cos\Big(\frac{3\pi}{4}+\text{x}\Big)\Big]$
$=\ -\Big[2\sin\frac{3\pi}{4}\sin\text{x}\Big]$ $[\because\ \cos(\text{A}-\text{B})-\cos(\text{A+B})=2\sin\text{A}\sin\text{B}]$
$=\ -2\sin\Big(\frac{\pi}{2}+\frac{\pi}{4}\Big)\sin\text{x}$
$=\ -2\cos\frac{\pi}{4}\sin\text{x}$ 
$=\ -2\times\frac{1}{\sqrt2}\times\sin\text{x}$
$=\ -\frac{\sqrt2\times\sqrt2}{\sqrt2}\sin\text{x}$
$=\ -\sqrt2\sin\text{x}$
$=\ \text{RHS}$
$\therefore\ \cos\Big(\frac{3\pi}{4}+\text{x}\Big)-\cos\Big(\frac{3\pi}{4}-\text{x}\Big)=-\sqrt2\sin\text{x}\ \text{Hence proved.}$
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Question 192 Marks
Prove that:
$\cos\frac{\pi}{12}-\sin\frac{\pi}{12}=\frac{1}{\sqrt2}$
Answer
$\text{LHS}=\cos\frac{\pi}{12}-\sin\frac{\pi}{12}$
Multiplying and dividing by $\sqrt2$ on LHS
$=\ \sqrt2\Big(\frac{1}{\sqrt2}\cos\frac{\pi}{12}-\frac{1}{\sqrt2}\sin\frac{\pi}{12}\Big)$
$=\ \sqrt2\Big(\sin\frac{\pi}{4}\cos\frac{\pi}{12}-\cos\frac{\pi}{4}\sin\frac{\pi}{12}\Big)$ $\Big[\because\ \frac{1}{\sqrt2}=\cos\frac{\pi}{4}=\sin\frac{\pi}{4}\Big]$
$=\ \sqrt2\Big(\sin\Big(\frac{\pi}{4}-\frac{\pi}{12}\Big)\Big)$ $[\because\ \sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$
$=\ \sqrt2\Big(\sin\frac{\pi}{6}\Big)$
$=\ \sqrt2\times\frac{1}{2}$
$=\ \frac{1}{\sqrt2}$
$=\ \text{RHS}$
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