To find the value of \(\cos(105^\circ)\), we break down \(105^\circ\) into the sum of two standard angles: \(105^\circ = 60^\circ + 45^\circ\). This allows us to use trigonometric identities.
The angle addition identity for cosine states: \(\cos(A + B) = \cos A \cos B - \sin A \sin B\)
Applying this identity with \(A = 60^\circ\) and \(B = 45^\circ\): \(\cos(105^\circ) = \cos(60^\circ + 45^\circ) = \cos(60^\circ)\cos(45^\circ) - \sin(60^\circ)\sin(45^\circ)\)
We use the known values for these standard angles:
Substitute these values into the expanded formula:
\(\cos(105^\circ) = \left(\frac{1}{2}\right) \left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{2}}{2}\right)\)
Simplify the terms:
\(\cos(105^\circ) = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}\)
Combine the terms over the common denominator:
\(\cos(105^\circ) = \frac{\sqrt{2} - \sqrt{6}}{4}\)
The calculated value is \(\frac{\sqrt{2} - \sqrt{6}}{4}\). Let's verify Option A: \(\frac{(1-\sqrt{3})}{2\sqrt{2}}\). To compare, we rationalize the denominator of Option A:
\(\frac{(1-\sqrt{3})}{2\sqrt{2}} = \frac{(1-\sqrt{3}) \times \sqrt{2}}{(2\sqrt{2}) \times \sqrt{2}} = \frac{\sqrt{2} - \sqrt{6}}{2 \times 2} = \frac{\sqrt{2} - \sqrt{6}}{4}\)
The calculated value matches Option A.
If \(cosec~\theta =\frac{29}{21}\) where 0 < θ < 90°, then what is the value of 4 sec θ + 4 tan θ?
The value of \(\cot \left( {cose{c^{ - 1}}\frac{5}{3} + {{\tan }^{ - 1}}\frac{2}{3}\;} \right)\)
The distance of the highest point on the graph of the function y = √3 cos x + sin x from the x-axis is:
If A = π / 6 and B = π / 3, then consider the following statements:
I. sin A + sin B = cos A + cos B
II. tan A + tan B = cot A + cot B
Which of the above statements is / are correct?
If \(\sin A = \frac{1}{{\sqrt 2 }}\) and \({\mathop{\rm Cos}\nolimits} B = \frac{{\sqrt 3 }}{2}\), then, find the value of (A + B)º.
A. 60º
B. 75º
C. 105º
D. 90º