# $\sin x+\sin y+\sin z=4\cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2}$ [duplicate]

If $$x+y+z=\pi$$ how to prove that $$\sin x+\sin y+\sin z=4\cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2}$$?

I got that $$\sin x+\sin y+\sin z=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}+\sin x\cos y+\sin y\cos x$$ and I don't know what to do next.

Can somebody help me, please?

## marked as duplicate by Martin R, drhab, Michael Rozenberg trigonometry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 2 at 13:23

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## 1 Answer

I hope you this helps with the question

=2*sin⁡((x+y)/2)* cos⁡((x-y)/2)+2*sin⁡(z/2)*cos⁡(z/2) =2*sin⁡((π-z)/2)* cos⁡((x-y)/2)+2*sin⁡((π-(x+y))/2) *cos⁡(z/2) =2*cos⁡(z/2)*cos⁡((x-y)/2)+2*cos⁡((x+y)/2)*cos⁡(z/2) =2*cos⁡(z/2)*[cos⁡((x-y)/2)+cos⁡((x+y)/2)] =2*cos⁡(z/2)*[2*cos⁡(x/2)*cos⁡(y/2)] =4*cos⁡(x/2)*cos⁡(y/2)*cos⁡(z/2)

• Please use MathJax to format your answers. As it stands this is not readable. – John Doe Apr 2 at 12:19
• It is not advisable to answer a question that has already been signaled as a duplicate (I should say multiplicate). – Jean Marie Apr 2 at 12:43