Trigonometric Functions - Test Papers

 CBSE Test Paper 01

CH-03 Trigonometric Functions

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1. In a triangle ABC, if A = 75o${75}^o$ and B = 45o${45}^o$ then b + 2 c is equal to
   1. a
   2. a + b + c
   3. 2 a
   4. 12(a+b+c)$\frac{1}{2}(a+b+c)$
2. cot θ$\theta$ = sin 2 θ(θ$\theta (\theta$ ≠$\ne$n π$\pi$ , n integer) if θ$\theta$ equals
   1. 90∘${90}^{\circ }$ only
   2. 45∘and60∘${45}^{\circ }and{60}^{\circ }$
   3. 45∘${45}^{\circ }$ only
   4. 45∘and90∘${45}^{\circ }and{90}^{\circ }$
3. If the angles of a triangle ABC are in A.P., then
   1. none of these
   2. c2=a2+b2$c^2=a^2+b^2$
   3. a2+c2−ac=b2$a^2+c^2-ac=b^2$
   4. c2=a2+b2+ab$c^2=a^2+b^2+ab$
4. In a triangle ABC, the line joining the circumcentre and the incentre is parallel to BC, then cos B + cos C =
   1. 32$\frac{3}{2}$
   2. 1
   3. 12$\frac{1}{2}$
   4. 34$\frac{3}{4}$
5. In a triangle ABC, AD is the median A to BC, then its length is equal to
   1. b+c2$\frac{b+c}{2}$
   2. b2+c2−a22−−−−−−−−−−√$\sqrt{b^2+c^2-\frac{a^2}{2}}$
   3. b2+c2−a22−−−−−−√$\sqrt{\frac{b^2+c^2-a^2}{2}}$
   4. 122(b2+c2)−a2−−−−−−−−−−−−√$\frac{1}{2}\sqrt{2(b^2+c^2)-a^2}$
6. Fill in the blanks:

   If sinθ$\theta$ + cosecθ$\theta$ = 2, then sin2θ$\theta$ + cosec2θ$\theta$ = ________.

7. Fill in the blanks:

   The general solution for sec x = sec (π+x$\pi +x$) is ________.

8. Evaluate: cos (- 870o)

9. Express as a product: cos 4x + cos 8x

10. Find the degree corresponding to the radian measure−2c$-2^c$

11. Prove sin26x-sin24x= sin 2x sin10 x

12. Solve: tan x + tan 2x + tan 3x = 0 

13. If axcosθ+bysinθ$\frac{ax}{\cos \theta }+\frac{by}{\sin \theta }$ = a2 - b2 and, axsinθcos2θ−bycosθsin2θ$\frac{ax\sin \theta }{{\cos }^2\theta }-\frac{by\cos \theta }{{\sin }^2\theta }$ = 0, prove that (ax)23+(by)23$(ax{)}^{\frac{2}{3}}+(by{)}^{\frac{2}{3}}$= (a2−b2)23$(a^2-b^2{)}^{\frac{2}{3}}$.

14. A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 66 m when it has traced out 45° at the centre, find the length of the rope.

15. Solve: 4 sinx cosx + 2 sinx + 2 cosx + 1 = 0.

CBSE Test Paper 01
CH-03 Trigonometric Functions

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Solution

1. (c) 2 a
   Explanation:

   Given A=75∘ and B=45∘$A={75}^{\circ }\text{ and }B={45}^{\circ }$
   then A+B+C=180∘$A+B+C={180}^{\circ }$
   ⇒75+45+C=180$\Rightarrow 75+45+C=180$
   ⇒$\Rightarrow$C=180-120=60
   From sin e formula
   asinA=bsinB=csinC=k(lct)$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k(\mathrm{lc}t)$
   ⇒asin75=bsin45=csin60=k$\Rightarrow \frac{a}{\sin 75}=\frac{b}{\sin 45}=\frac{c}{\sin 60}=k$
   Here
   a3√+122√=k,b12√=k,c3√2=k$\frac{a}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=k,\frac{b}{\frac{1}{\sqrt{2}}}=k,\frac{c}{\frac{\sqrt{3}}{2}}=k$Now,
   ⇒a=(3√+122√)k,b=k2√,c=3√k2$\Rightarrow a=\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)k,b=\frac{k}{\sqrt{2}},c=\frac{\sqrt{3}k}{2}$
   b+2c=k2√+2⋅3√k2=2k+23√k22√=2(3√+122√)k=2a$b+2c=\frac{k}{\sqrt{2}}+2\cdot \frac{\sqrt{3}k}{2}=\frac{2k+2\sqrt{3}k}{2\sqrt{2}}=2\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)k=2a$

    

2. (d) 45∘and90∘${45}^{\circ }and{90}^{\circ }$

   Explanation:
   sin2θ=cotθ$\sin 2\theta =\cot \theta$
   ⇒2sinθcosθ=sinθcosθ$\Rightarrow 2\sin \theta \cos \theta =\frac{\sin \theta }{\cos \theta }$
   ⇒cosθ(2sinθ−1sinθ)=0$\Rightarrow \cos \theta \left(2\sin \theta -\frac{1}{\sin \theta }\right)=0$
   ⇒−cosθ(1−2sin2θ)=0$\Rightarrow -\cos \theta \left(1-2{\sin }^2\theta \right)=0$
   ⇒−cosθ⋅cos2θ=0$\Rightarrow -\cos \theta \cdot \cos 2\theta =0$
   ⇒cosθ=cosπ2 or cos2θ=cosπ2$\Rightarrow \cos \theta =\cos \frac{\pi }{2}\text{ or }\cos 2\theta =\cos \frac{\pi }{2}$
   ⇒θ=π2,π4[∵θ≠nπ,n∈Z]$\Rightarrow \theta =\frac{\pi }{2},\frac{\pi }{4}[\because \theta \ne n\pi ,n\in Z]$

3. (c) a2+c2−ac=b2$a^2+c^2-ac=b^2$

   Explanation:

   Given angles of a triangle ABC are in A.P⇒A+C2=B$\Rightarrow \frac{A+C}{2}=B$ ⇒A+C=2B....(i)$\Rightarrow A+C=2B....(i)$

   But we have in a triangle A+B+C=180o⇒3B=180o⇒B=60o$$\begin{gathered} A+B+C={180}^{o \\ }\Rightarrow 3B={180}^{o \\ }\Rightarrow B={60}^{o \\ } \end{gathered}$$  [using(i)]
   Using Cosine Rule we have 

   b2=a2+c2−2ac.cosB⇒b2=a2+c2−2ac.cos60∘⇒b2=a2+c2−ac[∵cos60∘=12]$$\begin{gathered} b^2=a^2{+c}^2-2ac.cosB \\ \Rightarrow b^2=a^2{+c}^2-2ac.cos{60}^{\circ } \\ \Rightarrow b^2=a^2{+c}^2-ac\left[\because cos{60}^{\circ }=\frac{1}{2}\right] \end{gathered}$$

4. (b) 1
   Explanation:

   Using the given condition we have r = R cos A which gives r/R = cos A.hence Cos A + Cos B + cos C -1 = cos A which gives cos B + cos C = 1.

5. (d) 122(b2+c2)−a2−−−−−−−−−−−−√$\frac{1}{2}\sqrt{2(b^2+c^2)-a^2}$

   Explanation:

   Using the given information let the length of the median be d units, and let BD = DB =  m units. Using Apollonius theorem we have the following results.
   From triangle ADC , if the angle ADC is  θ$\theta$, then b2 = m2 + d2 - 2mdcos θ$\theta$. The angle ADB is the supplement of angle ADC and it will be - cos θ$\theta$. Hence from triangle ADB  we have b2 = m2 + d2+ 2mdcos θ$\theta$ . When we add both the results we have  b2 + c2 = 2 d2 + 2 m2 =  2 d2 + 2 (a2)2$(\frac{a}{2}{)}^2$ = 2 d2 + ( a22$\frac{a^2}{2}$).
   On simplifying we get the result.

6. 2

7. x = π2(2n−1)$\frac{\pi }{2}(2n-1)$

8. cos(−870o)=cos870o$cos(-{870}^o)=cos{870}^o$ [∵$\because$cos(−θ)=cos$cos(-\theta )=cos$ θ$\theta$]
   cos (π2×10−30)$\left(\frac{\pi }{2}\times 10-30\right)$ = ±$\pm$ cos 30o [∵$\because$ n is even]
   Now, α$\alpha$ =870o=$={870}^o=$ π2$\frac{\pi }{2}$ ×$\times$ 10−30o$10-{30}^o$
   It lies in II quadrant in which cos θ$\theta$ is negative.
   So, cos (- 870o) = - cos 30o = - 3√2$\frac{\sqrt{3}}{2}$

9. cos 4x + cos 8x
   =2cos(8x+4x2)cos(8x−4x2)$=2\cos \left(\frac{8x+4x}{2}\right)\cos \left(\frac{8x-4x}{2}\right)$ [∵cosC+cosD=2cosC+D2cosC−D2]$\left[\because \cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}\right]$
   = 2 cos 6x cos 2x

10. (−2)c=(180π×−2)∘=(18022×7×(−2))∘=(−114611)∘$(-2{)}^c={\left(\frac{180}{\pi }\times -2\right)}^{\circ }={\left(\frac{180}{22}\times 7\times (-2)\right)}^{\circ }={\left(-114\frac{6}{11}\right)}^{\circ }$
    
    =(−114∘(611×60)′)$=\left(-{114}^{\circ }{\left(\frac{6}{11}\times 60\right)}^{\mathrm{\prime }}\right)$

    =−[114∘(32811)′]=−[−114∘32′(811×60)′′]$=-\left[{114}^{\circ }{\left(32\frac{8}{11}\right)}^{\mathrm{\prime }}\right]=-\left[-{114}^{\circ }{32}^{\mathrm{\prime }}{\left(\frac{8}{11}\times 60\right)}^{\mathrm{\prime }\mathrm{\prime }}\right]$

    =−[114∘32′44′′]$=-\left[{114}^{\circ }{32}^{\mathrm{\prime }}{44}^{\mathrm{\prime }\mathrm{\prime }}\right]$

11. We have L.H.S. =sin26x−sin24x$={\sin }^{2}6x-{\sin }^{2}4x$
    =sin(6x+4x)⋅sin(6x−4x)$=\sin (6x+4x)\cdot \sin (6x-4x)$
    [∵sin2A−sin2B=sin(A+B)sin(A−B)]$[\because {\sin }^{2}A-{\sin }^{2}B=\sin (A+B)\sin (A-B)]$
    =sin10x⋅sin2x=R.H.S.$=\sin 10x\cdot \sin 2x=R.H.S.$

12. tan x + tan 2x + tan 3x = 0
    tan x + tan 2x + (tanx+tan2x)1−tanx⋅tan2x=0$\frac{(\tan x+\tan 2x)}{1-\tan x\cdot \tan 2x}=0$   [using tan 3x=(tanx+tan2x)1−tanx⋅tan2x$\frac{(\tan x+\tan 2x)}{1-\tan x\cdot \tan 2x}$]
    [tan x + tan 2x][1+11+tanxtan2x]=0$\left[1+\frac{1}{1+\tan x\tan 2x}\right]=0$
    [tan x + tan 2x][2 - tan x tan 2x] = 0
    tan x = -tan 2x or tan x tan 2x = 2
    x=nπ−2x or tanx⋅2tanx1+tan2x=2$\mathrm{x}=\mathrm{n}\pi -2\mathrm{x}\text{ or }\tan \mathrm{x}\cdot \frac{2\tan \mathrm{x}}{1+{\tan }^2\mathrm{x}}=2$
    3x=nπ or 2tan2x=2−2tan2x$3\mathrm{x}=\mathrm{n}\pi \text{ or }2{\mathrm{t}\mathrm{a}\mathrm{n}}^2\mathrm{x}=2-2{\tan }^2\mathrm{x}$
    3x=nπ or 4tan2x=2$3\mathrm{x}=\mathrm{n}\pi \text{ or }4{\tan }^2\mathrm{x}=2$
    x=nπ3 or tan 2x=12$\mathrm{x}=\frac{\mathrm{n}\pi }{3}{\text{ or tan }}^2\mathrm{x}=\frac{1}{2}$
    x=nπ3 or x=mπ+tan−1(12√),n,m∈Z$\mathrm{x}=\frac{\mathrm{n}\pi }{3}\text{ or }\mathrm{x}=\mathrm{m}\pi +{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right),n,m\in Z$

13. Given,
    axsinθcos2θ−bycosθsin2θ$\frac{ax\sin \theta }{{\cos }^2\theta }-\frac{by\cos \theta }{{\sin }^2\theta }$ =0$=0$
    ⇒$\Rightarrow$ax sin3$axsi{n}^3$ θ$\theta$ - by cos3$byco{s}^3$ θ$\theta$ =0$=0$
    ⇒$\Rightarrow$ sin3θby$\frac{{\sin }^3\theta }{by}$ = cos3θax$\frac{{\cos }^3\theta }{ax}$ 
    ⇒$\Rightarrow$(sin3θby)2/3${\left(\frac{{\sin }^3\theta }{by}\right)}^{2/3}$ = (cos3θax)2/3${\left(\frac{{\cos }^3\theta }{ax}\right)}^{2/3}$
    ⇒$\Rightarrow$ sin2θ(by)2/3$\frac{{\sin }^2\theta }{(by{)}^{2/3}}$ = cos2θ(ax)2/3$\frac{{\cos }^2\theta }{(ax{)}^{2/3}}$
    ⇒$\Rightarrow$ sin2θ(by)2/3$\frac{{\sin }^2\theta }{(by{)}^{2/3}}$ = cos2θ(ax)2/3$\frac{{\cos }^2\theta }{(ax{)}^{2/3}}$ = sin2θ+cos2θ(by)2/3+(ax)2/3$\frac{{\sin }^2\theta +{\cos }^2\theta }{(by{)}^{2/3}+(ax{)}^{2/3}}$ [Using ratio and proportions]
    ⇒$\Rightarrow$ sin2θ(by)2/3$\frac{{\sin }^2\theta }{(by{)}^{2/3}}$ = cos2θ(ax)2/3$\frac{{\cos }^2\theta }{(ax{)}^{2/3}}$ = sin2θ+cos2θ(by)2/3+(ax)2/3$\frac{{\sin }^2\theta +{\cos }^2\theta }{(by{)}^{2/3}+(ax{)}^{2/3}}$= 1(by)2/3+(ax)2/3$\frac{1}{(by{)}^{2/3}+(ax{)}^{2/3}}$
    ⇒$\Rightarrow$ sin2$si{n}^2$ θ$\theta$ = (by)2/3(ax)2/3+(by)2/3$\frac{(by{)}^{2/3}}{(ax{)}^{2/3}+(by{)}^{2/3}}$ and, cos$cos$2θ$\theta$ = (ax)2/3(ax)2/3+(by)2/3$\frac{(ax{)}^{2/3}}{(ax{)}^{2/3}+(by{)}^{2/3}}$
    ⇒$\Rightarrow$ sin$sin$ θ$\theta$ = (by)1/3(ax)2/3+(by)2/3√$\frac{(by{)}^{1/3}}{\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}}$ and, cos$cos$ θ$\theta$ = (ax)1/3(ax)2/3+(by)2/3√$\frac{(ax{)}^{1/3}}{\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}}$
    Substituting the values in axcosθ+bysinθ$\frac{ax}{\cos \theta }+\frac{by}{\sin \theta }$ =a2−b2$=a^2-b^2$, 
    ⇒(ax)2/3(ax)2/3+(by)2/3−−−−−−−−−−−−−√+(by)2/3(ax)2/3+(by)2/3−−−−−−−−−−−−−√$\Rightarrow (ax{)}^{2/3}\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}+(by{)}^{2/3}\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}$ =a2−b2$=a^2-b^2$
    ⇒$\Rightarrow$ {(ax)2/3+(by)2/3−−−−−−−−−−−−−√}{(ax)2/3+(by)2/3}$\left\{\sqrt{(ax{)}^{2/3}+(by{)}^{2/3}}\right\}\left\{(ax{)}^{2/3}+(by{)}^{2/3}\right\}$ =a2−b2$=a^2-b^2$
    ⇒$\Rightarrow$ {(ax)2/3+(by)2/3}3/2${\left\{(ax{)}^{2/3}+(by{)}^{2/3}\right\}}^{3/2}$ = a2 - b2 ⇒$\Rightarrow$ (ax)2/3+(by)2/3$(ax{)}^{2/3}+(by{)}^{2/3}$ = (a2−b2)2/3${\left(a^2-b^2\right)}^{2/3}$

14. Here PA = PB = r
    arc AB = 66 m and θ=45∘$\theta ={45}^{\circ }$
    Now θ=45∘=(45×π180)C=πC4$\theta ={45}^{\circ }={\left(45\times \frac{\pi }{180}\right)}^C=\frac{{\pi }^C}{4}$
    
    We know that
    θ=1r$\theta =\frac{1}{r}$
    ∴π4=66r⇒r=66×422×7=84m$\therefore \frac{\pi }{4}=\frac{66}{r}\Rightarrow r=\frac{66\times 4}{22}\times 7=84m$

15. 4sinxcosx+2sinx+2cosx+1=0$4\sin x\cos x+2\sin x+2\cos x+1=0$
    ⇒$\Rightarrow$ 2sin x(2cos x+1)+1(2cos x+1)=0$2sinx(2cosx+1)+1(2cosx+1)=0$
    ⇒$\Rightarrow$ (2sin x+1)(2cos x+1)=0$(2sinx+1)(2cosx+1)=0$
    ⇒$\Rightarrow$ 2sin x+1=0$2sinx+1=0$
    or 2cos x+1=0$2cosx+1=0$
    ⇒$\Rightarrow$ sinx=−$sinx=-$ 12$\frac{1}{2}$
    or cosx=−$cosx=-$ 12$\frac{1}{2}$
    Now, if sin x = - 12$\frac{1}{2}$
    ⇒$\Rightarrow$ sin x = sin (−π6)$\left(-\frac{\pi }{6}\right)$
    ∴$\therefore$ The general solution of this equation is
    x = ​​​​nπ+(−1)n(−π6)$n\pi +(-1{)}^n\left(-\frac{\pi }{6}\right)$= nπ+(−1)n+1(π6)$n\pi +(-1{)}^{n+1}\left(\frac{\pi }{6}\right)$
    ⇒$\Rightarrow$ x = π[n+(−1)n+16]$\pi \left[n+\frac{(-1{)}^{n+1}}{6}\right]$ ...(i)
    and if cos x = −12$\frac{-1}{2}$
    ⇒$\Rightarrow$ cos x = cos (π−π3)$\left(\pi -\frac{\pi }{3}\right)$ = cos 2π3$\frac{2\pi }{3}$
    The general solution of this equation is
    x = 2nπ±2π3$2n\pi \pm \frac{2\pi }{3}$
    ⇒$\Rightarrow$ x = 2π(n±13)$2\pi \left(n\pm \frac{1}{3}\right)$ ... (ii)
    From Eqs. (i) and (ii), we have x = π[n+(−1)n+16]$\pi \left[n+\frac{(-1{)}^{n+1}}{6}\right]$ or  2π(n±13)$2\pi \left(n\pm \frac{1}{3}\right)$ where n ∈$\in$ Z
    These are the required solutions.