Question

Cos a = 3/5 Найдите sin a/2, cos a/2, tg a/2

Answer 1

$cos\alpha =\frac{3}{5}\\\\sin^2\frac{\alpha }{2}=\frac{1-cos\alpha }{2}=\frac{1-\frac{3}{5}}{2}=\frac{2}{5\cdot 2}=\frac{1}{5}\; \; ,\; \; sin\frac{\alpha }{2}=\pm \frac{1}{\sqrt5}=\pm \frac{\sqrt5}{5}\\\\cos^2\frac{\alpha }{2}=\frac{1+cos\alpha }{2}=\frac{1+\frac{3}{5}}{2}=\frac{8}{5\cdot 2}=\frac{4}{5}\; \; ,\; \; cos\frac{\alpha }{2}=\pm \frac{2}{\sqrt5}=\pm \frac{2\sqrt5}{5}\\\\tg\frac{\alpha }{2}=\pm \sqrt{\frac{1-cos\alpha }{1+cos\alpha }}=\pm \sqrt{\frac{1-\frac{3}{5}}{1+\frac{3}{5}}}=\pm \sqrt{\frac{2}{8}}=\pm \frac{1}{2}$

Answer 2

To find the values of sin(a/2), cos(a/2), and tan(a/2) based on the given value of cos(a) = 3/5, we can use the half-angle identities. These identities are derived from the double-angle identities and provide a way to express the trigonometric functions of an angle in terms of the trigonometric functions of half that angle.

Let's start by finding sin(a/2):

sin(a/2) = ±√[(1 - cos(a))/2]

Since we are given cos(a) = 3/5, we can substitute this value into the equation:

sin(a/2) = ±√[(1 - 3/5)/2] = ±√[(2/5)/2] = ±√(2/10) = ±√(1/5) = ±(1/√5) = ±(1/√5) * (√5/√5) = ±√5/5

Therefore, sin(a/2) can have two possible values: √5/5 or -√5/5.

Next, let's find cos(a/2):

cos(a/2) = ±√[(1 + cos(a))/2]

Substituting the value of cos(a) = 3/5:

cos(a/2) = ±√[(1 + 3/5)/2] = ±√[(8/5)/2] = ±√(8/10) = ±√(4/5) = ±(2/√5) = ±(2/√5) * (√5/√5) = ±2√5/5

So, cos(a/2) can have two possible values: 2√5/5 or -2√5/5.

Finally, let's find tan(a/2):

tan(a/2) = sin(a/2) / cos(a/2)

Substituting the values we obtained earlier:

tan(a/2) = (√5/5) / (2√5/5) = √5/5 * 5/2√5 = 1/2

Hence, tan(a/2) = 1/2.

To summarize: sin(a/2) can be either √5/5 or -√5/5. cos(a/2) can be either 2√5/5 or -2√5/5. tan(a/2) is always 1/2.