![If z(1)= sqrt(2)(cos ""(pi)/(4)+ isin""(pi)/(4))" and "z(2)= sqrt(3)(cos ""( pi)/(3)+ isin"" (pi)/(3)), then |z(1)z(2)| is If z(1)= sqrt(2)(cos ""(pi)/(4)+ isin""(pi)/(4))" and "z(2)= sqrt(3)(cos ""( pi)/(3)+ isin"" (pi)/(3)), then |z(1)z(2)| is](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/433384974_web.png)
If z(1)= sqrt(2)(cos ""(pi)/(4)+ isin""(pi)/(4))" and "z(2)= sqrt(3)(cos ""( pi)/(3)+ isin"" (pi)/(3)), then |z(1)z(2)| is
![दिखाइए की ` (i) cosx +sin x =sqrt(2) cos (x-( pi)/(4)) =sqrt(2) sin (x+ ( pi )/(4)) ` ` (ii) sqrt(3) - YouTube दिखाइए की ` (i) cosx +sin x =sqrt(2) cos (x-( pi)/(4)) =sqrt(2) sin (x+ ( pi )/(4)) ` ` (ii) sqrt(3) - YouTube](https://i.ytimg.com/vi/gR-Vrcs9R9M/maxresdefault.jpg?sqp=-oaymwEmCIAKENAF8quKqQMa8AEB-AH-CYAC0AWKAgwIABABGGUgZShlMA8=&rs=AOn4CLC7CmUvZ9H43ro_F6yGMShfyRP9Vg)
दिखाइए की ` (i) cosx +sin x =sqrt(2) cos (x-( pi)/(4)) =sqrt(2) sin (x+ ( pi )/(4)) ` ` (ii) sqrt(3) - YouTube
![trigonometry - Solve the trignometric equation $2\sin (3x + \pi/4) =\sqrt {1+8\sin2x\cos^2x}$ - Mathematics Stack Exchange trigonometry - Solve the trignometric equation $2\sin (3x + \pi/4) =\sqrt {1+8\sin2x\cos^2x}$ - Mathematics Stack Exchange](https://i.stack.imgur.com/CI2PU.jpg)
trigonometry - Solve the trignometric equation $2\sin (3x + \pi/4) =\sqrt {1+8\sin2x\cos^2x}$ - Mathematics Stack Exchange
![See answer: Given w= sqrt2 (cos ( pi / 4 ) + i sin ( pi / 4 ) ) and z = 2 ( cos ( pi / 2 ) + I sin ( pi /2 ) - Brainly.com See answer: Given w= sqrt2 (cos ( pi / 4 ) + i sin ( pi / 4 ) ) and z = 2 ( cos ( pi / 2 ) + I sin ( pi /2 ) - Brainly.com](https://us-static.z-dn.net/files/dc6/198c687dd11e464b6b8fbba814ab0bd1.jpg)
See answer: Given w= sqrt2 (cos ( pi / 4 ) + i sin ( pi / 4 ) ) and z = 2 ( cos ( pi / 2 ) + I sin ( pi /2 ) - Brainly.com
![trigonometry - Simpler Derivation of $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, - Mathematics Stack Exchange trigonometry - Simpler Derivation of $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, - Mathematics Stack Exchange](https://i.stack.imgur.com/gwI1X.jpg)
trigonometry - Simpler Derivation of $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, - Mathematics Stack Exchange
![when applying the Laplace transform why is the -pi/4 in cos(t-pi/4) ignored ? Thank you. : r/askmath when applying the Laplace transform why is the -pi/4 in cos(t-pi/4) ignored ? Thank you. : r/askmath](https://i.redd.it/g603dvpfv1441.png)
when applying the Laplace transform why is the -pi/4 in cos(t-pi/4) ignored ? Thank you. : r/askmath
![By Using the Properties of Definite Integrals, Evaluate the Integrals `Int_0^(Pi/2) Sqrt(Sinx)/(Sqrt(Sinx) + Sqrt(Cos X)) Dx - Mathematics | Shaalaa.com By Using the Properties of Definite Integrals, Evaluate the Integrals `Int_0^(Pi/2) Sqrt(Sinx)/(Sqrt(Sinx) + Sqrt(Cos X)) Dx - Mathematics | Shaalaa.com](https://www.shaalaa.com/images/_4:dcdc367888c840b7ba97d3b40009964c.png)