Question

Sin^2A=
it's from trigonometry
​

Answer 1

Answer:

2sinAcosA

Step-by-step explanation:

sinAcosA+cosAsinA

2(sinAcosA)

2sinA√[1-cos^2A)

Answer 2

❒Given to find the value of :-

sin2A

❒ Solution:-

sin2A can be wrtitten as sin(A+A)

sin(A+A) in the form of sin(A+B)

As we know that ,

$\green\bigstar\green{\boxed{sin(A+B) = sinAcosB + sinB cosA}}$

Now we shall expand sin(A+A)

sin(A+ A) = sinA cosA + sinA cosA

sin(A+A) = 2 sinA cosA

So, the value of sin2A = 2sinAcosA

❒❒❒❒❒❒❒❒❒❒❒❒❒❒❒

_____________________

❒Know more formulae:-

❒Multiple angles :-

$sin2A = 2sinAcosA$

$sin2A = \dfrac{2tanA}{1 + tan {}^{2}A }$

$cos2A = cos {}^{2} A - sin { }^{2} A$

$cos2A= 2cos {}^{2} A - 1$

$cos2A = 1 - 2in {}^{2} A$

$cos2A = \dfrac{1 - tan {}^{2} A}{1 + tan {}^{2}A }$

$tan2A = \dfrac{2tanA}{1 - tan {}^{2}A }$

$cot2A = \dfrac{cot {}^{2} A - 1}{2cotA}$

$sin3A = 3sinA - 4sin {}^{3} A$

$cos3A = 4cos {}^{3} A - 3cosA$

$tan3A = \dfrac{3tanA - tan {}^{3}A }{1 - 3tan {}^{2} A}$

❒Compound angles

$sin(A+B)= sinAcosB + sinBcosA$

$sin(A-B) = sinAcosB- sinBcosA$

$cos(A+B) = cosAcosB - sinAsinB$

$cos(A-B) = cosAcosB + sinAsinB$

$tan(A+B) = \dfrac{tanA+tanB}{1-tanAtanB}$

$tan(A-B) = \dfrac{tanA-tanB}{1+tanAtanB}$

$cot(A+B) = \dfrac{cotB cotA -1}{cotB + cotA}$

$cot(A-B) = \dfrac{cotB cotA + 1}{cotB-cotA}$