Question

If X and Y are acute angles such that sinX=4/5 and cosY =12/13. calculate without using tables.
(a) sin(X-Y) (b) cos(X-Y)
Also show that: cosX+cos(X+30°)+sin(X+60°)= 3/5(1+√3).

Answer 1

Step-by-step explanation:

For sin X = 4/5 , => p =4 h = 5 , => b = 3

For sinY = 12/13 , => p= 12 , h = 13 , => b = 5

sin(X - Y) = sin X cos Y - cos X sin Y

= 4/5 × 5/13 - 3/5 × 12/13 = (20 - 36)/65

= - 16/65

cos(X - Y) = cos X cos Y + sin X sin Y

= 3/5 × 5/13 + 4/5 × 12/13 = 16/65 + 48/65

= 64/65

cos X + cos( X+ 30°) + sin ( X + 60°)

= cos X + cos X cos 30° - sin X sin 30° + sin X cos 60° + cos X sin 60° = cos X + cos X × √3/2 - sin X ×1/2 + sin X × 1/2 + cos X × √3/2

= cos X ( 1 + √3) = 3/5 ( 1 + √3 ).

Answer 2

Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies.

For $sin X = 4/5 , = > p =4 h = 5 , = > b = 3$

For $sinY = 12/13 , = > p= 12 , h = 13 , = > b = 5$

$sin(X - Y) = sin X cos Y - cos X sin Y$

$= 4/5 \times 5/13 - 3/5 \times 12/13 = (20 - 36)/65$

$= - 16/65$

$cos(X - Y) = cos X cos Y + sin X sin Y$

$= 3/5 \times 5/13 + 4/5 \times 12/13 = 16/65 + 48/65$

$= 64/65$

$cos X + cos( X+ 30^\circ) + sin ( X + 60^\circ)$

$= cos X + cos X cos 30^\circ - sin X sin 30^\circ + sin X cos 60^\circ + cos X sin 60^\circ = cos X + cos X \times \sqrt{3}/2 - sin X \times 1/2 + sin X \times 1/2 + cos X \times \sqrt{3} /2$

$= cos X ( 1 + \sqrt{3}) = 3/5 ( 1 + \sqrt{3} )$

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