Trigonometry Questions

Solution : The value of sin 30 degrees is \(1\over 2\). Proof : Consider an equilateral triangle ABC with each side of length of 2a. Each angle of \(\Delta\) ABC is of 60 degrees. Let AD be the perpendicular from A on BC. \(\therefore\)   AD is the bisector of \(\angle\) A and D is the […]

Solution : The value of Cosec 0 degrees is equal to \(\infty\) or not defined. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B

Solution : The value of Sec 0 degrees is equal to 1. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B and finally

Solution : The value of Cot 0 degrees is equal to \(\infty\) or not defined. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B

Solution : The value of Tan 0 degrees is equal to 0. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B and finally when

Solution : The value of Cos 0 degrees is equal to 1. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B and finally when

Solution : The value of sin 0 degrees is equal to 0. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B and finally

Solution : We have 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\) = 1 + \(1\over sqrt{2}\)\((cos\theta + cos\theta)\) + \(\sqrt{2}\)\((cos\theta + cos\theta)\) = 1 + \(({1\over \sqrt{2}} + \sqrt{2})\) + \((cos\theta + cos\theta)\) = 1 + \(({1\over \sqrt{2}} + \sqrt{2})\).\(\sqrt{2}cos(\theta – {pi\over 4})\) \(\therefore\)   Maximum Value = 1 + \(({1\over \sqrt{2}}

Solution : The expression = (sin78 – sin42) – (sin66 – sin6) = 2cos(60)sin(18) – 2cos36.sin30 = sin18 – cos36 = \(({\sqrt{5} – 1\over 4})\) – \(({\sqrt{5} + 1\over 4})\) = -\(1\over 2\) Similar Questions Find the maximum value of 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\) If A + B

Solution : cos2A + cos2B + cos2C = 2cos(A+B)cos(A-B) + cos2C = 2cos(\(3\pi\over 2\) – C)cos(A-B) + cos2C  \(\because\)  A + B + C = \(3\pi\over 2\) = -2sinC cos(A-B) + 1 – 2\(sin^2C\) = 1 – 2sinC[cos(A-B)+sinC] = 1 – 2sinC[cos(A-B) + sin(\(3\pi\over 2\)-(A+B))] = 1 – 2sinC[cos(A-B)-cos(A+B)] = 1 – 4sinA sinB sinC