Sine & Cosine

Geometry Level 2

If $\sqrt{3}\cos \theta-\sin \theta=\frac{1}{3}$ and $0 < \theta < \frac{\pi}{2},$ the value of $\sqrt{3}\sin \theta+\cos \theta$ can be expressed as $\frac{\sqrt{a}}{3}$ . What is the value of $a?$

20 25 30 35

1 solution

Naved Husain
Feb 24, 2014

Cos30cos(theta)-sin30sin(theta)=1/6 Cos(30+theta)=1/6 (base=1 and hypotenuse=6 therefore perpendicular =sqrt(35)

The second term can be written as Sin(30+theta)=sqrt(35)/6 Therefore a=35

Sqrt(3) cos(theta)-sin(theta)=1/3 =>2 sin(theta-pi/3)=1/3 =>theta=sin^-1(1/3)+pi/3=tan^-1(1/sqrt(35))...(@) From the given conditions Divding the equations we get [Sqrt(3)cos(theta)-sin(theta)]/[Sqrt(3) sin(theta)+cos(theta)]=1/sqrt(a) Dividing numerator and denominator by cos(theta), [tan(pi/3)-tan(theta)]/[1+tan(pi/3)tan(theta) ]=1/sqrt(a) =>tan(theta-pi/3)=1/sqrt(a) =>tan(tan^-1(1/sqrt(35))=1/sqrt(a).........by (@) =>a=35

Yohenba Soibam - 1 year, 11 months ago

Why, on a page filled with expressions a sin (theta) + b cos (theta) is this expression a cos (theta) - b sin(theta) without any explanation of how to put the given expression into the form that the formula applies to. I don’t understand how Sqrt (3) cos (theta) - sin (theta) = 1/3 becomes 2 sin (theta.pi/3) = 1/3

Lindsay Fischer - 1 year, 1 month ago

Sine & Cosine

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