Question

1. cos32°+cos28° 2. sin63°-sin27° 3. cos3/4pi+sin11/12pi

Answer 1

$cos32+cos28=2cos \frac{32+28}{2}cos \frac{32-28}{2}=2cos30cos2=2 *\frac{ \sqrt{3} }{2}cos2= \\ \sqrt{3}cos2; \\ sin63-sin27=2sin \frac{63-27}{2}cos \frac{63+27}{2}=2sin18cos45=2* \frac{ \sqrt{2} }{2}sin18= \\ \sqrt{2}sin18; \\ cos \frac{3\pi}{4}+sin \frac{11\pi}{12}= sin( \frac{\pi}{2}- \frac{3\pi}{4}) +sin \frac{11\pi}{12}= \\ 2sin \frac{ \frac{\pi}{2}- \frac{3\pi}{4}+ \frac{11\pi}{12}}{2}cos \frac{ \frac{\pi}{2}- \frac{3\pi}{4}- \frac{11\pi}{12} }{2}=2sin \frac{2\pi}{3}cos \frac{7\pi}{6}=$
$=2* \frac{ \sqrt{3} }{2} *(- \frac{ \sqrt{3}}{2} )=- \frac{3}{2} =-1,5$

Answer 2

To solve these trigonometric expressions, we'll use the trigonometric identities and a calculator. Here are the solutions:

1. cos(32°) + cos(28°):

Using the cosine addition formula: cos(A + B) = cos(A)cos(B) - sin(A)sin(B)

cos(32°) + cos(28°) = cos(32°) + cos(28°) = cos(32°)cos(28°) - sin(32°)sin(28°)

Using a calculator, we find: cos(32°) ≈ 0.848 cos(28°) ≈ 0.882 sin(32°) ≈ 0.533 sin(28°) ≈ 0.469

Substituting these values: cos(32°) + cos(28°) ≈ 0.848 * 0.882 - 0.533 * 0.469 ≈ 0.747 - 0.25 ≈ 0.497

Therefore, cos(32°) + cos(28°) ≈ 0.497.

2. sin(63°) - sin(27°):

Using the sine subtraction formula: sin(A - B) = sin(A)cos(B) - cos(A)sin(B)

sin(63°) - sin(27°) = sin(63°) - sin(27°) = sin(63°)cos(27°) - cos(63°)sin(27°)

Using a calculator, we find: sin(63°) ≈ 0.891 sin(27°) ≈ 0.454 cos(63°) ≈ 0.454 cos(27°) ≈ 0.891

Substituting these values: sin(63°) - sin(27°) ≈ 0.891 * 0.891 - 0.454 * 0.454 ≈ 0.794 - 0.206 ≈ 0.588

Therefore, sin(63°) - sin(27°) ≈ 0.588.

3. cos(3/4π) + sin(11/12π):

Let's convert the angles to degrees:

3/4π radians = (3/4) * 180°/π ≈ 135° 11/12π radians = (11/12) * 180°/π ≈ 165°

Using a calculator, we find: cos(135°) ≈ -0.707 sin(165°) ≈ -0.258

Substituting these values: cos(3/4π) + sin(11/12π) ≈ -0.707 + (-0.258) ≈ -0.965

Therefore, cos(3/4π) + sin(11/12π) ≈ -0.965.